Functions, polynomials and the binomial. Attnom Academy

M5.1.1 Chapter 1

Domain, range and inverses

Can I put any number into a function, and can every number come out?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A Yes. A function accepts every number and can give every number.
You picked A. Read on. At the end, the chapter comes back to your guess.
B No. A function has a set of inputs it accepts, and its outputs may fill only part of the number line.
You picked B. Read on. At the end, the chapter comes back to your guess.
C Inputs are limited, but the outputs can always be any number.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Find the domain and range of a function
Say when a function has an inverse
Sketch a function and its inverse, mirrored in y = x

A function is a rule that gives exactly one output for each allowed input. The set of allowed inputs is the domain. The set of outputs that the rule really gives is the range.

Take f(x) = x². If we allow only the inputs −2, −1, 1 and 2, the domain is that list. The arrows show the range is 1 and 4.

f(x) = x² on the domain −2, −1, 1, 2. The range is 1 and 4.

Two inputs may share an output, as −2 and 2 do here. That is allowed. A rule like this is called many to one. What a function never does is give one input two outputs.

Some papers write f : x → 2x + 1, with a small bar at the back of the arrow, and read it "f maps x to 2x + 1". This means f(x) = 2x + 1.

Unless a question says otherwise, the domain is every real number for which the rule works. A real number is any number on the number line, including fractions and roots such as √2.

We write a domain or range in two ways. Set notation uses braces: {x : x ≥ 2} reads "the set of x such that x is at least 2". Interval notation uses brackets: [2, ∞).

A square bracket includes the end. A round bracket leaves it out. The sign ∞ means "goes on without end". It is not a number, so it always has a round bracket. So [−1, 3] means −1 ≤ x ≤ 3, and (−1, 3] means −1 < x ≤ 3.

The three lines show [−1, 3], then (−1, 3], then [2, ∞). A filled circle means the end is included. An open circle means it is left out.

Three things can stop an input. You cannot divide by zero. You cannot take the square root of a negative number and stay with real numbers. And a question may restrict the domain, for example "for 0 ≤ x ≤ 3".

To find the range, ask which outputs are possible. A sketch helps: read the lowest and highest heights the curve reaches. A restricted domain gives its own range.

The graph stops at the end points (0, 1) and (3, 7), so it shows only the domain asked for.

Worked example

f(x) = 2x + 1 for 0 ≤ x ≤ 3. Write the domain and find the range.

The domain is given: 0 ≤ x ≤ 3, which is [0, 3].
The line y = 2x + 1 rises as x grows, so the lowest output comes from the smallest input.
Smallest input: f(0) = 2 × 0 + 1 = 1.
Largest input: f(3) = 2 × 3 + 1 = 7.
A straight line has no gaps, so every height from 1 to 7 is reached.
The range is 1 ≤ y ≤ 7, which is [1, 7].

Answer: The range is [1, 7].

Now a square root. Its inside must be 0 or more, and the root sign gives the positive root, so the output is never below 0.

The curve starts at x = 2. Nothing is drawn to the left, and nothing is below the height 1.

Worked example

f(x) = √(x − 2) + 1. Find the largest possible domain, and the range.

A real square root needs the inside to be 0 or more: x − 2 ≥ 0.
Add 2 to both sides: x ≥ 2. The domain is [2, ∞).
The root sign gives the positive root, so √(x − 2) ≥ 0, and it equals 0 when x = 2.
Add 1 to the root: f(x) = √(x − 2) + 1 ≥ 1.
At the start, f(2) = √0 + 1 = 1. As x grows the outputs grow without end, with no gaps.
The range is y ≥ 1, which is [1, ∞).

Answer: The domain is [2, ∞) and the range is [1, ∞).

A fraction cannot have a bottom of zero. That removes one input, and often one output as well.

g(x) = 1/(x − 3). The curve has no point on the dashed line x = 3, and it never touches the x axis.

Worked example

g(x) = 1/(x − 3). Find the domain and the range.

The bottom must not be zero. x − 3 = 0 when x = 3, so x ≠ 3.
The domain is every real number except 3.
For the range, write y = 1/(x − 3) and try to solve for x.
Multiply by (x − 3): y(x − 3) = 1. If y ≠ 0, divide by y: x − 3 = 1/y.
So x = 3 + 1/y. This gives an input for every y except y = 0, where 1/y has no meaning.
Test it: g(4) = 1/(4 − 3) = 1, and g(2) = 1/(2 − 3) = −1.
The outputs fall on both sides of 0, but 0 needs 1/(x − 3) = 0, and a fraction with top 1 is never 0. The range is every real number except 0.

Answer: The domain is x ≠ 3 and the range is y ≠ 0.

Now the inverse. A function is one to one if different inputs always give different outputs. Only a one to one function has an inverse function f⁻¹, because the inverse must take each output back to a single input.

The horizontal line test shows it. If every horizontal line cuts the graph at most once, the function is one to one.

The dashed line y = 4 cuts y = x² twice. So x² on all real numbers is not one to one.

We can mend this by restricting the domain. Keep only x ≥ 0. Now each output has one input, and the inverse is f⁻¹(x) = √x.

The inverse swaps inputs and outputs. So the domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f. If (a, b) is on the graph of f, then (b, a) is on the graph of f⁻¹. Swapping x and y reflects the graph in the line y = x.

f(x) = x² for x ≥ 0 and its inverse f⁻¹(x) = √x. Each curve is the mirror image of the other in the dashed line y = x.

To find an inverse: write y = f(x), solve for x, then write x for the input again. Check which root to take from the domain of f.

f(x) = (x − 2)² + 1 and f⁻¹(x) = 2 + √(x − 1), from the example below. The graphs are mirror images in y = x, and the start (2, 1) of f swaps to the start (1, 2) of f⁻¹.

Worked example

f(x) = x² − 4x + 5 for x ≥ 2. Write f(x) as (x − p)² + q. Find the range of f. Find f⁻¹(x), and give its domain and range.

Complete the square: x² − 4x + 5 = (x − 2)² − 4 + 5 = (x − 2)² + 1.
For x ≥ 2 the bracket x − 2 is 0 or more, and (x − 2)² grows as x grows.
The smallest output is f(2) = 0 + 1 = 1, and the outputs grow without end. The range is y ≥ 1.
Each output now comes from only one input on x ≥ 2, so f is one to one and has an inverse.
Write y = (x − 2)² + 1, so (x − 2)² = y − 1.
Take the square root. On this domain x − 2 is not negative, so take the positive root: x − 2 = √(y − 1).
So x = 2 + √(y − 1). Write x for the input: f⁻¹(x) = 2 + √(x − 1).
The domain of f⁻¹ is the range of f, which is x ≥ 1. The range of f⁻¹ is the domain of f, which is y ≥ 2.

Answer: f⁻¹(x) = 2 + √(x − 1), with domain x ≥ 1 and range y ≥ 2.

Two more words help when a question asks for them. The codomain is the set the outputs are allowed to lie in, which a question may state. The range is the part of the codomain that is really reached. A function is onto if its range is the whole codomain.

Take f(x) = x² from the real numbers to the real numbers. It is not onto: the range is y ≥ 0, so no input gives −1. But f(x) = x² from the real numbers to [0, ∞) is onto, because every y ≥ 0 equals x² for x = √y. A function that is one to one and onto is a bijection, and every bijection has an inverse.

Now two functions in a row. In fg(x) = f(g(x)), the function g goes first. So fg exists only for inputs x where g(x) is in the domain of f. This can shrink the domain of fg.

Worked example

f(x) = √(x − 2) and g(x) = 3 − x. Find fg(x) and its domain.

fg(x) = f(g(x)). Put g(x) = 3 − x into f: fg(x) = √((3 − x) − 2) = √(1 − x).
The domain of g is all real numbers, so any x can go into g.
But f needs its input to be 2 or more, because √(x − 2) needs x − 2 ≥ 0. So we need g(x) ≥ 2: 3 − x ≥ 2.
Take 3 from both sides: −x ≥ −1. Multiply by −1 and turn the sign round: x ≤ 1.
Check with x = 1: g(1) = 2, and f(2) = 0, so fg(1) = √0 = 0. And x = 2 fails: g(2) = 1, and f(1) = √(−1) does not exist.
So fg(x) = √(1 − x), with domain x ≤ 1.

Answer: fg(x) = √(1 − x), with domain x ≤ 1.

Try it

Fold the graph along y = x

You need: squared paper, a pencil and a ruler.

Draw axes from 0 to 6 on both sides. Draw the line y = 2x + 1 and mark the points (0, 1), (1, 3) and (2, 5).
Swap the two numbers of each point and mark the new points (1, 0), (3, 1) and (5, 2).
Draw the dashed line y = x. Draw a line through the new points. It is the graph of f⁻¹(x) = (x − 1)/2.
Fold the paper along the dashed line. Hold it to the light.

What you will see: Each point of the first line lands on its swapped point. The two lines are mirror images in y = x.

If you cannot try it, this is what happens.

A wrong idea many people have

f⁻¹(x) means 1/f(x).

Why it feels right: A raised −1 on a number gives a reciprocal: 2⁻¹ is 1/2. The same mark on f seems to say the same thing.

The test that breaks it: Take f(x) = x + 2. Then f(3) = 5, so the inverse takes 5 back to 3: f⁻¹(5) = 3. The reciprocal would give 1/f(5) = 1/7, which is a different number.

What is true: For a function, f⁻¹ is the inverse function, which undoes f. The reciprocal is written 1/f(x) or (f(x))⁻¹, and it is a different thing.

A wrong idea many people have

Every function has an inverse.

Why it feels right: For straight line functions such as 2x + 1 we can always undo the steps, so it seems any function can be undone.

The test that breaks it: Take f(x) = x² on all real numbers. f(3) = 9 and f(−3) = 9. An inverse would have to send 9 back to both 3 and −3, and a function may not give two outputs.

What is true: Only a one to one function has an inverse. Restrict the domain first, for example to x ≥ 0, and the inverse exists on that domain.

A wrong idea many people have

The range of x² − 4x + 5 is y ≥ 5, because 5 is the last number.

Why it feels right: For y = x² + c the lowest point is c, so the last number seems to give the lowest output.

The test that breaks it: The −4x term moves the lowest point. f(2) = 4 − 8 + 5 = 1, and 1 is below 5.

What is true: Complete the square. In (x − p)² + q the lowest value is q, so here it is 1. Check the domain too, because it may cut off the lowest point.

Key words

function: a rule that gives exactly one output for each allowed input.
domain: the set of inputs a function allows.
range: the set of outputs a function really gives.
real number: any number on the number line, including fractions and roots.
one to one: different inputs always give different outputs.
inverse function: the function f⁻¹ that takes each output back to its input. It exists only if f is one to one.
restricting the domain: allowing only some inputs, so that a function becomes one to one.

Back to your guess

The answer is B: No. A function has a set of inputs it accepts, and its outputs may fill only part of the number line.

Take f(x) = √x. You cannot put in −4, because a negative number has no real square root. And the output is never negative either. The inputs allowed are called the domain, and the outputs reached are called the range. Both can be only part of the number line.

Keep it

Copy this into your exercise book.

Domain: inputs allowed. Range: outputs reached. Only a one to one function has an inverse; restrict the domain to get one. f⁻¹ swaps x and y, so reflect the graph in y = x.

Draw: A curve and its inverse, with the dashed line y = x between them and a pair of mirror points.

M5.1.2 Chapter 2

The modulus function

How far is a number from zero, when I do not care which way I walk?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A The distance from −5 to 0 is −5.
You picked A. Read on. At the end, the chapter comes back to your guess.
B The distance from −5 to 0 is 5.
You picked B. Read on. At the end, the chapter comes back to your guess.
C The distance from −5 to 0 depends on which way you walk.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Sketch y = |2x − 3|
Solve |x − 2| < 5

The modulus of a number is its distance from 0 on the number line. It is written with two bars: |x|. Some books call it the absolute value.

Both −5 and 5 are 5 steps from 0. So |−5| = 5 and |5| = 5.

A distance is never negative, so |x| ≥ 0 for every x, and |x| = 0 only when x = 0. We define the modulus in two pieces.

|x| = x when x ≥ 0. |x| = −x when x < 0. When x is negative, −x is positive, so |−5| = −(−5) = 5.

More generally, |a − b| is the distance between the numbers a and b. It is the same as |b − a|, because the distance does not depend on the direction.

So |x − 2| < 5 says: the distance from x to 2 is less than 5. On the number line that is every number within 5 steps of 2, on either side.

The numbers within 5 steps of 2: from 5 below 2 to 5 above 2. The ends −3 and 7 are not included, so their circles are open.

This gives a rule. For b > 0: |x − a| < b means a − b < x < a + b. And |x − a| > b means x < a − b or x > a + b. If b is negative, |x − a| < b has no solution, because a modulus is never negative.

Worked example

Solve |x − 2| < 5.

The distance from x to 2 must be less than 5.
So x − 2 lies between −5 and 5: −5 < x − 2 < 5.
Add 2 to every part. The left end becomes −5 + 2 = −3.
The right end becomes 5 + 2 = 7. So −3 < x < 7.

Answer: −3 < x < 7

Now the graph. Take y = 2x − 3. For x ≥ 1.5 the value 2x − 3 is 0 or more, so |2x − 3| = 2x − 3. For x < 1.5 it is negative, so |2x − 3| = −(2x − 3) = 3 − 2x.

So the graph of y = |2x − 3| is the line y = 2x − 3 where it is on or above the x axis. Where the line is below the axis, the graph is its reflection in the x axis. The graph has a corner on the axis.

Only the part of y = 2x − 3 that lies below the x axis is dashed. It is folded up into the left arm of the solid V.

Worked example

Sketch y = |2x − 3|. Mark where it meets the axes.

Find where 2x − 3 = 0: 2x = 3, so x = 1.5. The graph touches the x axis at (1.5, 0).
Put x = 0: |2 × 0 − 3| = |−3| = 3. The graph meets the y axis at (0, 3).
For x ≥ 1.5 the graph is the line y = 2x − 3, with gradient 2.
For x < 1.5 the graph is the line y = 3 − 2x, with gradient −2.
The two pieces meet in a V shape, with its corner at (1.5, 0). The graph is never below the x axis.

Answer: A V with its corner at (1.5, 0), meeting the y axis at (0, 3).

The graph also solves inequalities. To solve |2x − 3| < 5, draw the line y = 5 and find where the V is below it.

The V is below the dashed line y = 5 between x = −1 and x = 4.

Worked example

Solve |2x − 3| < 5.

The inside 2x − 3 must lie between −5 and 5: −5 < 2x − 3 < 5.
Add 3 to every part: −2 < 2x < 8.
Divide every part by 2: −1 < x < 4.
Check with the graph: the V is below y = 5 between x = −1 and x = 4.
Check one number inside, x = 0: |2 × 0 − 3| = 3, and 3 < 5. Check one outside, x = 5: |2 × 5 − 3| = 7, and 7 is not less than 5.

Answer: −1 < x < 4

When the right side of an equation is not a plain number, a modulus equation needs more care. We split into two cases, from the two pieces of the definition. Then we must check every answer in the original equation.

g(x) = |x − 1| and f(x) = 2x + 1 meet once, at (0, 1). The other answer from the algebra, x = −2, has the line at height −3, where the V can never be.

Worked example

Solve |x − 1| = 2x + 1.

Case 1: x ≥ 1. Then |x − 1| = x − 1, so x − 1 = 2x + 1.
Take x from both sides: −1 = x + 1, so x = −2.
But −2 is not ≥ 1, so case 1 gives no answer.
Case 2: x < 1. Then |x − 1| = −(x − 1) = 1 − x, so 1 − x = 2x + 1.
Add x to both sides and take 1 away: 0 = 3x, so x = 0.
And 0 < 1, so case 2 gives x = 0.
Check in the original: |0 − 1| = 1 and 2 × 0 + 1 = 1. They agree.
Check the rejected number too: |−2 − 1| = 3, but 2 × (−2) + 1 = −3. A modulus is never −3, so −2 fails.

Answer: x = 0 only.

The same fold works for a curve. For y = |x² − 4|, draw y = x² − 4, which dips below the x axis between −2 and 2. Fold that dip up. The result has a corner at each of x = −2 and x = 2, where the curve meets the axis.

f(x) = |x² − 4| and the dashed part of g(x) = x² − 4 that is folded up.

A different change is y = f(|x|). Put |x| in place of x. For x ≥ 0 nothing changes. For x < 0, |x| = −x, so the graph copies the right half in a mirror image in the y axis. So y = |x| − 2 is a V with its corner at (0, −2).

Two moduli can be compared by squaring. For any real numbers a and b, |a| < |b| exactly when a² < b². This is true because |a|² = a², and squaring keeps the order of non-negative numbers.

Worked example

Solve |2x − 1| < |x + 3|.

Square both sides. This is allowed because both sides are moduli, which are 0 or more: (2x − 1)² < (x + 3)².
Expand: 4x² − 4x + 1 < x² + 6x + 9.
Collect everything on the left: 3x² − 10x − 8 < 0.
Factorise: (3x + 2)(x − 4) < 0.
The product is negative between the two roots. The roots come from 3x + 2 = 0 and x − 4 = 0, which give x = −2/3 and x = 4.
So −2/3 < x < 4.
Check with x = 0: |−1| = 1 and |3| = 3, and 1 < 3. Check with x = 5: |9| = 9 and |8| = 8, and 9 is not less than 8.

Answer: −2/3 < x < 4

Try it

Fold up the part below the axis

You need: squared paper, a pencil.

Make a table of x and 2x − 3 for x from −1 to 4.
Add a third row for |2x − 3|. Drop every minus sign, because a modulus is never negative.
Plot both rows as two graphs on one grid. The first is a straight line, the second is a V.
Fold the paper along the x axis. The part of the line below the axis lands on the V.

What you will see: Where 2x − 3 is already 0 or more, the two rows agree. Where it is negative, the modulus row shows the same number without its sign.

If you cannot try it, this is what happens.

A wrong idea many people have

The solution of |x − 2| < 5 is x < 7.

Why it feels right: Taking the bars away gives x − 2 < 5, and that gives x < 7. It seems that only the right end matters.

The test that breaks it: Try x = −10. Then |−10 − 2| = 12, which is not less than 5. But −10 is less than 7, so x < 7 lets in numbers that fail.

What is true: The bars mean two limits. Write −5 < x − 2 < 5, which gives −3 < x < 7. A modulus below a number is between two numbers.

A wrong idea many people have

|a + b| = |a| + |b|.

Why it feels right: A modulus looks like a sign you can remove from each piece, like opening a bracket.

The test that breaks it: Take a = 3 and b = −5. Then |a + b| = |−2| = 2. But |a| + |b| = 3 + 5 = 8.

What is true: In general |a + b| ≤ |a| + |b|. Equality holds only when a and b have the same sign, or one is 0. A modulus does not split over addition.

A wrong idea many people have

Solving |x − 1| = 2x + 1 gives two answers, x = −2 and x = 0, and both are right.

Why it feels right: A modulus equation often has two answers, so two answers from the algebra look right.

The test that breaks it: Put x = −2 back: |−2 − 1| = 3, but 2 × (−2) + 1 = −3. The two sides are not equal.

What is true: When the right side has x in it, check every answer in the original equation. The right side must not be negative, because a modulus never is.

Key words

modulus: the distance of a number from 0, written |x|. It is x when x ≥ 0 and −x when x < 0. Also called the absolute value.
modulus function: the function f(x) = |x|, or a function with a modulus in its rule, such as |2x − 3|.

Back to your guess

The answer is B: The distance from −5 to 0 is 5.

A distance is never negative. From −5 to 0 is 5 steps, and from 5 to 0 is also 5 steps. The modulus of a number is this distance from 0, so |−5| = 5 and |5| = 5.

Keep it

Copy this into your exercise book.

|x| is the distance from 0. |x − a| < b means a − b < x < a + b. y = |f(x)| folds the part below the x axis upwards. Check every answer to |…| = an expression in x.

Draw: A line crossing the x axis, with the part below it folded up into a V.

M5.1.3 Chapter 3

Dividing polynomials

Can I divide x³ + 2x² − 5x − 6 by x + 1 the way I divide 86 by 7?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A No. Only numbers can be divided.
You picked A. Read on. At the end, the chapter comes back to your guess.
B Yes. The same steps work, one power of x at a time.
You picked B. Read on. At the end, the chapter comes back to your guess.
C Yes, but only if the answer has no remainder.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Divide x³ + 2x² − 5x − 6 by x + 1
Find a remainder quickly with the remainder theorem

A polynomial in x is a sum of terms. Each term is a number times a power of x, and the power is a whole number: 0, 1, 2, 3, and so on. For example 4x³ − 2x² + x + 7 is a polynomial.

The number in front of a power is its coefficient. The highest power of x is the degree of the polynomial. The term with no x is the constant term. So 1/x and √x are not polynomials.

The polynomial 4x³ − 2x² + x + 7 has degree 3. Its constant term is 7, and the coefficient of x is 1.

With numbers, 17 ÷ 5 = 3 remainder 2 means 17 = 5 × 3 + 2. The remainder is smaller than 5. With polynomials we say the same thing, with the word "smaller" meaning "lower degree".

If f(x) is divided by d(x), with quotient q(x) and remainder r(x), then f(x) = d(x) × q(x) + r(x), and r(x) has a lower degree than d(x). Dividing by x − a gives a remainder with degree 0: just a number.

x² + 5x + 7 divided by x + 2. Each power of x has its own column, so the terms stay in line.

Worked example

Divide x² + 5x + 7 by x + 2.

Divide the first terms: x² ÷ x = x. Write x above the x column.
Multiply the divisor by x: x(x + 2) = x² + 2x. Write it under the dividend.
Subtract. The x² terms cancel, and 5x − 2x = 3x.
Bring down the 7. The new line is 3x + 7.
Divide the first terms: 3x ÷ x = 3. Write + 3 above.
Multiply: 3(x + 2) = 3x + 6. Subtract: (3x + 7) − (3x + 6) = 1.
The remainder 1 has degree 0, which is lower than the degree 1 of the divisor. Stop.

Answer: The quotient is x + 3 and the remainder is 1.

To check, multiply back: (x + 2)(x + 3) + 1 = x² + 5x + 6 + 1 = x² + 5x + 7. This is the dividend, so the division is right.

Now a cubic. The steps repeat once for each power. Watch each line build up in the picture below.

Draw it first

Divide x³ + 2x² − 5x − 6 by x + 1

1 Divide x³ by x to get x². Multiply x²(x + 1) = x³ + x², subtract, and bring down to get x² − 5x.
2 Divide x² by x to get x. Multiply x(x + 1) = x² + x, subtract, and bring down to get −6x − 6.
3 Divide −6x by x to get −6. Multiply −6(x + 1) = −6x − 6, subtract, and the remainder is 0.

Worked example

Divide x³ + 2x² − 5x − 6 by x + 1.

Divide the first terms: x³ ÷ x = x². Write x² above the x² column.
Multiply: x²(x + 1) = x³ + x². Subtract from the top: (x³ + 2x²) − (x³ + x²) = x². Bring down −5x to get x² − 5x.
Divide the first terms: x² ÷ x = x. Write + x above.
Multiply: x(x + 1) = x² + x. Subtract: (x² − 5x) − (x² + x) = −6x. Bring down −6 to get −6x − 6.
Divide the first terms. Write − 6 above, because −6x ÷ x = −6.
Multiply: −6(x + 1) = −6x − 6. Subtract: (−6x − 6) − (−6x − 6) = 0.
The remainder is 0, so x + 1 divides exactly. So x³ + 2x² − 5x − 6 = (x + 1)(x² + x − 6).

Answer: The quotient is x² + x − 6 and the remainder is 0.

If a power of x is missing, write it in with coefficient 0. This keeps the columns in line. So x³ − 7x + 6 is written x³ + 0x² − 7x + 6.

x³ − 7x + 6 divided by x − 1. The missing x² term is written as + 0x², so that −7x stays in the x column.

Worked example

Divide x³ − 4x² + x + 9 by x − 2. Write the result as f(x) = (x − 2) × q(x) + r.

Divide x³ ÷ x = x². Multiply: x²(x − 2) = x³ − 2x². Subtract: (x³ − 4x²) − (x³ − 2x²) = −2x². Bring down x: −2x² + x.
Divide −2x² ÷ x = −2x. Multiply: −2x(x − 2) = −2x² + 4x. Subtract: (−2x² + x) − (−2x² + 4x) = −3x. Bring down 9: −3x + 9.
Divide −3x ÷ x = −3. Multiply: −3(x − 2) = −3x + 6. Subtract: (−3x + 9) − (−3x + 6) = 3.
The quotient is x² − 2x − 3 and the remainder is 3.
So x³ − 4x² + x + 9 = (x − 2)(x² − 2x − 3) + 3.
Check at x = 2. The bracket (x − 2) is 0, so the right side is 3. And the left side is 8 − 16 + 2 + 9 = 3.

Answer: The quotient is x² − 2x − 3 and the remainder is 3.

The check in the last step is the key. When x = 2 the divisor x − 2 is zero, so the whole product vanishes and only the remainder is left. The remainder equals the value of the polynomial at x = 2.

f(x) = x³ − 4x² + x + 9. At x = 2 the height of the curve is 3, the remainder when f(x) is divided by x − 2.

The remainder theorem says: when a polynomial f(x) is divided by x − a, the remainder is f(a).

Proof. Divide by x − a. The remainder has degree less than 1, so it is a number r. Then f(x) = (x − a) q(x) + r for every x. Put x = a. The bracket is 0, so f(a) = 0 × q(a) + r = r.

Take care with the sign. The divisor x + 2 is x − (−2), so a = −2 and the remainder is f(−2). In general, the divisor x + b gives f(−b), and the divisor 2x − 1 gives f(1/2).

Worked example

Find the remainder when f(x) = 2x³ − 5x² + x + 7 is divided by x + 2.

Write the divisor as x − (−2). So a = −2.
By the remainder theorem the remainder is f(−2).
Work out each term: 2 × (−2)³ = 2 × (−8) = −16, and −5 × (−2)² = −5 × 4 = −20.
The other terms: x = −2, and the constant is 7.
Add: f(−2) = −16 − 20 − 2 + 7 = −31.

Answer: The remainder is −31.

Worked example

The polynomial f(x) = x³ + px² − 4x + 3 leaves remainder 11 when divided by x − 2. Find p. Then find the remainder when f(x) is divided by x + 1.

The divisor is x − 2, so a = 2 and the remainder is f(2).
f(2) = 2³ + p × 2² − 4 × 2 + 3 = 8 + 4p − 8 + 3 = 4p + 3.
The remainder is 11, so 4p + 3 = 11. Then 4p = 8, so p = 2.
So f(x) = x³ + 2x² − 4x + 3.
The divisor x + 1 is x − (−1), so the remainder is f(−1).
f(−1) = (−1)³ + 2 × (−1)² − 4 × (−1) + 3 = −1 + 2 + 4 + 3 = 8.

Answer: p = 2, and the second remainder is 8.

For a divisor such as 2x − 1, write f(x) = (2x − 1)q(x) + r. At x = 1/2 the bracket 2x − 1 is 0, so r = f(1/2). In general, the divisor px − q gives the remainder f(q/p), where q/p is the value of x that makes the divisor 0.

Worked example

Find the remainder when f(x) = 4x³ − 2x + 5 is divided by 2x − 1.

The divisor 2x − 1 is 0 when x = 1/2. So the remainder is f(1/2).
f(1/2) = 4 × (1/2)³ − 2 × (1/2) + 5.
(1/2)³ = 1/8, so 4 × 1/8 = 1/2. And 2 × 1/2 = 1.
So f(1/2) = 1/2 − 1 + 5 = 9/2.

Answer: The remainder is 9/2.

A divisor can have degree 2 as well. The remainder then has degree 1 or less, so it can hold an x term. The long division is the same.

x³ + 3x² − 2x + 4 divided by x² + 1. The quotient is x + 3. The remainder −3x + 1 has degree 1, which is less than 2.

Worked example

Divide x³ + 3x² − 2x + 4 by x² + 1.

Divide the first terms: x³ ÷ x² = x. Write x above the x column.
Multiply: x(x² + 1) = x³ + x. Subtract: (x³ + 3x²) − (x³ + 0x²) = 3x². The x column gives −2x − x = −3x. So the line is 3x² − 3x.
Divide the first terms: 3x² ÷ x² = 3. Write + 3 above.
Multiply: 3(x² + 1) = 3x² + 3. Subtract: (3x² − 3x + 4) − (3x² + 3) = −3x + 1.
The remainder −3x + 1 has degree 1, which is less than the degree 2 of the divisor. Stop.

Answer: The quotient is x + 3 and the remainder is −3x + 1.

Try it

Multiply back to check a division

You need: paper, a pencil.

Take the result of the first worked example: quotient x + 3, divisor x + 2, remainder 1.
Draw a grid with x and + 2 across, and x and + 3 down. Multiply each pair of terms.
Add the four products and collect like terms.
Add the remainder 1. Compare with the dividend x² + 5x + 7.

What you will see: The grid gives x² + 5x + 6. Adding the remainder 1 gives x² + 5x + 7, the dividend. A division is right when dividend = divisor × quotient + remainder.

If you cannot try it, this is what happens.

A wrong idea many people have

To find the remainder when dividing by x + 2, work out f(2).

Why it feels right: The number 2 is the one written in the divisor, so it seems the one to put in.

The test that breaks it: Take f(x) = x + 2 itself. It divides exactly by x + 2, so the remainder is 0. But f(2) = 4, and f(−2) = 0.

What is true: The divisor x + 2 is x − (−2). Put in x = −2. Write the divisor as x − a first, and then the number is a.

A wrong idea many people have

A missing power of x can be left out of the long division.

Why it feels right: The term is not there, so it seems there is nothing to write.

The test that breaks it: In x³ − 7x + 6 the −7x would then stand in the x² column. The first subtraction would mix an x term with an x² term, and the answer would be wrong.

What is true: Write the missing power with coefficient 0: x³ + 0x² − 7x + 6. Each column then holds one power.

A wrong idea many people have

To subtract (x² − 5x) − (x² + x) you get −5x + x = −4x.

Why it feels right: The first term cancels, and then we carry on adding what is left.

The test that breaks it: Multiply back: (x + 1)(x² + x − 4) = x³ + 2x² − 3x − 4. This is not the dividend x³ + 2x² − 5x − 6, so the division is wrong.

What is true: Subtract every term of the line below: change the sign of each term. (x² − 5x) − (x² + x) = −5x − x = −6x.

Key words

polynomial: a sum of terms, each a number times a whole number power of x.
coefficient: the number in front of a power of x.
degree: the highest power of x in a polynomial.
quotient: the result of a division, without the remainder.
remainder: what is left when the division stops. It has a lower degree than the divisor.
remainder theorem: the remainder of f(x) divided by x − a is f(a).

Back to your guess

The answer is B: Yes. The same steps work, one power of x at a time.

The steps for long division of numbers work for polynomials too. At each step you divide the first term by the first term of the divisor, multiply back and subtract. A remainder is allowed, as it is with numbers, and it can be found without any division at all.

Keep it

Copy this into your exercise book.

Divide first terms, multiply back, subtract, bring down. dividend = divisor × quotient + remainder. Remainder of f(x) ÷ (x − a) is f(a). Divisor x + 2 means a = −2.

Draw: A long division in columns, one column for each power of x, with 0 as a place holder.

M5.1.4 Chapter 4

The factor theorem

How can one number tell me that x − 2 divides a polynomial exactly?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A It cannot. You must do the whole division to find out.
You picked A. Read on. At the end, the chapter comes back to your guess.
B Only when the polynomial is a quadratic.
You picked B. Read on. At the end, the chapter comes back to your guess.
C Put x = 2 into the polynomial. If the answer is 0, then x − 2 is a factor.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Use f(a) = 0 to show that (x − a) is a factor
Factorise a cubic fully and sketch its graph

A factor of a polynomial divides it exactly, with no remainder. By the remainder theorem, the remainder on dividing f(x) by x − a is f(a). So the remainder is 0 exactly when f(a) = 0.

The factor theorem says: x − a is a factor of f(x) if and only if f(a) = 0. "If and only if" means it works both ways.

Proof. Suppose f(a) = 0. Then the remainder on dividing by x − a is f(a) = 0, so x − a divides exactly and is a factor. Now suppose x − a is a factor. Then f(x) = (x − a) q(x) for some q(x). Put x = a: f(a) = 0 × q(a) = 0.

The theorem has a form for brackets like 2x − 1. The bracket ax − b is a factor of f(x) if and only if f(b/a) = 0. The reason: ax − b = a(x − b/a), so it is a factor exactly when x − b/a is a factor, and that happens exactly when f(b/a) = 0. For 2x − 1, test f(1/2).

A value a with f(a) = 0 is a root of the equation f(x) = 0. It is also called a zero of f. On the graph, a root is a point where the curve meets the x axis. So the factor theorem links three things: a root a, a factor x − a, and a crossing point (a, 0).

How do we find a root? For a polynomial with whole number coefficients, any whole number root divides the constant term. For x³ − 3x² − 4x + 12 the constant is 12, so try the divisors of 12, positive and negative.

Why does this work? If a is a whole number root, then f(a) = 0 gives a³ + … + c = 0, where every term except c has a as a factor. So a must divide c. A polynomial may also have roots that are not whole numbers, and then this list will not find them.

Trying values in f(x) = x³ − 3x² − 4x + 12. The first two fail. The third gives 0, so x − 2 is a factor.

Once one factor is known, divide it out. What is left is a quadratic, which we factorise as before. Three steps: find a root, divide, factorise the rest.

Dividing x³ − 3x² − 4x + 12 by x − 2. The remainder is 0, as the factor theorem promised.

Worked example

Show that x − 2 is a factor of f(x) = x³ − 3x² − 4x + 12. Then factorise f(x) completely.

Work out f(2) = 2³ − 3 × 2² − 4 × 2 + 12 = 8 − 12 − 8 + 12 = 0.
Since f(2) = 0, the factor theorem says x − 2 is a factor.
Divide f(x) by x − 2. The quotient is x² − x − 6 (see the picture) and the remainder is 0.
Factorise the quadratic: find two numbers with product −6 and sum −1. They are −3 and 2.
So x² − x − 6 = (x − 3)(x + 2).
So f(x) = (x − 2)(x − 3)(x + 2). Written with the roots in order: (x + 2)(x − 2)(x − 3).
Check by expanding the first two brackets: (x + 2)(x − 2) = x² − 4, and (x² − 4)(x − 3) = x³ − 3x² − 4x + 12. This is f(x).

Answer: (x + 2)(x − 2)(x − 3)

Now the sketch. Three facts fix the shape of a cubic. First, the roots: the curve meets the x axis at each one. Second, the y intercept: put x = 0 and read off the constant term. Third, the ends: for x³ with a positive coefficient, the curve starts low on the left and ends high on the right.

f(x) = (x + 2)(x − 2)(x − 3). The dots on the x axis are the roots −2, 2 and 3, written under them. The curve meets the y axis at (0, 12).

Worked example

Sketch y = x³ − 3x² − 4x + 12, marking where it meets the axes.

From the example above, y = (x + 2)(x − 2)(x − 3). The curve meets the x axis at x = −2, x = 2 and x = 3.
Put x = 0: y = 0³ − 3 × 0² − 4 × 0 + 12 = 12. The curve meets the y axis at (0, 12).
The coefficient of x³ is 1, which is positive. So y is large and negative for large negative x, and large and positive for large positive x.
Between the roots the sign changes. The curve is above the axis between −2 and 2, and below the axis between 2 and 3.
Draw a smooth curve through the roots, with no corners: up through −2, over a hump, down through 2, under the axis, and up through 3.

Answer: The graph meets the x axis at (−2, 0), (2, 0) and (3, 0), and the y axis at (0, 12).

A cubic does not always cross three times. A repeated factor such as (x − 1)² gives a repeated root. The curve touches the axis there and turns back, instead of crossing.

f(x) = (x − 1)²(x + 2) = x³ − 3x + 2. The curve crosses the x axis at (−2, 0) and only touches it at (1, 0).

Worked example

f(x) = 2x³ + x² − 13x + 6. Show that x − 2 is a factor. Factorise f(x) completely and solve f(x) = 0.

f(2) = 2 × 8 + 4 − 26 + 6 = 16 + 4 − 26 + 6 = 0.
Since f(2) = 0, the factor theorem says x − 2 is a factor.
Divide f(x) by x − 2. The quotient is 2x² + 5x − 3 (see the picture).
Factorise 2x² + 5x − 3. Find two numbers with product 2 × (−3) = −6 and sum 5. They are 6 and −1.
Split the middle term: 2x² + 6x − x − 3 = 2x(x + 3) − (x + 3) = (2x − 1)(x + 3).
So f(x) = (x − 2)(2x − 1)(x + 3).
f(x) = 0 when x − 2 = 0, or 2x − 1 = 0, or x + 3 = 0. So x = 2, x = 1/2 or x = −3.

Answer: x = −3, x = 1/2 or x = 2.

The division in the example above. The first term of the dividend is 2x³, so the first quotient term is 2x².

Sometimes a letter is unknown, and the factor theorem finds it. The fact that x − a is a factor gives one equation, f(a) = 0, in the unknown.

f(x) = x³ − 2x² − x − 6, the cubic of the next example. It crosses the x axis only once, at (3, 0).

Worked example

f(x) = x³ − 2x² + kx − 6, and x − 3 is a factor. Find k. Then factorise f(x) as far as possible, and say how many real roots f(x) = 0 has.

Since x − 3 is a factor, f(3) = 0.
f(3) = 27 − 18 + 3k − 6 = 3 + 3k.
So 3 + 3k = 0, which gives k = −1.
So f(x) = x³ − 2x² − x − 6. Divide by x − 3: the quotient is x² + x + 2 and the remainder is 0.
So f(x) = (x − 3)(x² + x + 2).
Try to factorise x² + x + 2. The discriminant is b² − 4ac = 1 − 8 = −7, which is negative, so it has no real roots.
The only real root of f(x) = 0 is x = 3. The cubic crosses the x axis once.

Answer: k = −1, and the only real root is x = 3.

Try it

Find where a cubic crosses

You need: squared paper, a pencil and a ruler.

Make a table of f(x) = x³ − 3x² − 4x + 12 for x from −3 to 4.
Look down the f(x) column. Mark every x where f(x) is 0.
Plot the points and join them with a smooth curve.
Check that the curve meets the x axis at the values you marked.

What you will see: The values f(x) = 0 happen at x = −2, 2 and 3. These are the roots, and each gives a factor x + 2, x − 2 and x − 3.

If you cannot try it, this is what happens.

A wrong idea many people have

If f(2) = 0, then x + 2 is a factor.

Why it feels right: The number 2 appears, so a plus sign seems as good as a minus sign.

The test that breaks it: Take f(x) = x² − 5x + 6. Then f(2) = 4 − 10 + 6 = 0, so x − 2 is a factor. But f(−2) = 4 + 10 + 6 = 20, which is not 0, so x + 2 is not a factor.

What is true: f(a) = 0 gives the factor x − a. The sign flips. For a factor x + 2, test f(−2).

A wrong idea many people have

If f(1) and f(−1) are not 0, then f(x) has no factor of the form x − a.

Why it feels right: For many exercises the roots are 1, −1, 2 or −2, so a failed first try seems to say there are none.

The test that breaks it: Take g(x) = 2x³ − x² + 2x − 1. Then g(1) = 2 and g(−1) = −6, neither is 0. But g(1/2) = 1/4 − 1/4 + 1 − 1 = 0, so 2x − 1 is a factor.

What is true: When the first coefficient is not 1, roots can be fractions p/q, where p divides the constant term and q divides the first coefficient. Try those as well.

A wrong idea many people have

A cubic must cross the x axis three times.

Why it feels right: A cubic has degree 3, and it seems to have three roots, one for each power.

The test that breaks it: Take f(x) = (x − 3)(x² + x + 2). The quadratic factor has no real roots, so the curve crosses the axis once, at x = 3. And (x − 1)²(x + 2) meets the axis at only two points.

What is true: A cubic meets the x axis at one, two or three points. At two points it crosses at one and touches at the other. It always meets the axis at least once, because its ends go in opposite directions.

Key words

factor: a polynomial that divides another exactly, with no remainder.
factor theorem: x − a is a factor of f(x) if and only if f(a) = 0.
root: a value a with f(a) = 0. On the graph it is where the curve meets the x axis.
repeated root: a root from a repeated factor such as (x − 1)². The curve touches the axis there.

Back to your guess

The answer is C: Put x = 2 into the polynomial. If the answer is 0, then x − 2 is a factor.

The remainder theorem says the remainder on dividing by x − 2 is f(2). A division with no remainder is exact, so x − 2 is a factor exactly when f(2) = 0. One sum replaces a whole long division.

Keep it

Copy this into your exercise book.

f(a) = 0 means x − a is a factor. Try divisors of the constant term, divide, then factorise the quadratic. Roots are crossings. A repeated root touches.

Draw: A cubic crossing the x axis at three marked roots, with its y intercept marked.

M5.1.5 Chapter 5

Partial fractions

Can I split (3x + 5)/((x + 1)(x + 2)) into fractions with one bracket each on the bottom?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A No. A fraction with a product on the bottom cannot be split.
You picked A. Read on. At the end, the chapter comes back to your guess.
B Yes, by splitting the top: 3x/((x + 1)(x + 2)) + 5/((x + 1)(x + 2)).
You picked B. Read on. At the end, the chapter comes back to your guess.
C Yes. It becomes one fraction for each bracket on the bottom, each with a plain number on top.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Split (3x + 5)/((x + 1)(x + 2)) into two simpler fractions
Split a fraction with a repeated factor on the bottom

To add 2/(x + 1) and 1/(x + 2), we write them over the common bottom (x + 1)(x + 2). The top becomes 2(x + 2) + 1(x + 1) = 3x + 5. So 2/(x + 1) + 1/(x + 2) = (3x + 5)/((x + 1)(x + 2)).

Partial fractions run this backwards. We start with the single fraction (3x + 5)/((x + 1)(x + 2)) and split it into the parts it came from. The parts are plainer, and later they can be expanded as a series (the last chapter of this book) and, in calculus, integrated one at a time.

The method works when the top has a lower degree than the bottom. When the top is as big as the bottom, or bigger, divide first. We come to that at the end.

For a bottom made of different brackets, write one fraction for each bracket, with a letter on top. Here (3x + 5)/((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2). We must find the numbers A and B.

A statement that is true for every value of x is called an identity, and it is written with the sign ≡. So 3x + 5 ≡ A(x + 2) + B(x + 1) means the two sides are equal for every x. That is why we may put in any value of x we like.

There are two ways. The first is to clear the bottoms, and then equate coefficients. The second is to put in clever values of x, which is the cover up shortcut. Both give the same answer, so each checks the other.

The cover up for (3x + 5)/((x + 1)(x + 2)). Cover one bracket, put in the x that makes it zero, and work out the rest.

Worked example

Split (3x + 5)/((x + 1)(x + 2)) into partial fractions.

Write (3x + 5)/((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2).
Multiply both sides by (x + 1)(x + 2). The bottoms cancel: 3x + 5 = A(x + 2) + B(x + 1).
Expand the right side: 3x + 5 = (A + B)x + (2A + B).
Equate the coefficients of x: A + B = 3. Equate the constants: 2A + B = 5.
Subtract the first equation from the second: A = 2. Then B = 3 − 2 = 1.
So (3x + 5)/((x + 1)(x + 2)) = 2/(x + 1) + 1/(x + 2).
Check with the cover up. Cover x + 1 and put x = −1 in the rest: (3(−1) + 5)/(−1 + 2) = 2/1 = 2, which is A. Cover x + 2 and put x = −2: (3(−2) + 5)/(−2 + 1) = (−1)/(−1) = 1, which is B.
Check with a number. At x = 0 the left side is 5/2. The right side is 2/1 + 1/2 = 5/2.

Answer: 2/(x + 1) + 1/(x + 2)

The cover up works because putting in x = −1 makes the term with B vanish, since B is multiplied by (x + 1), which is 0. What stays is A times the other bracket. So A is the top divided by the other bracket, with x = −1.

Now a repeated factor. For a bottom with (x + 1)², one fraction is not enough. We need a fraction for the bracket once and a fraction for the bracket squared. The form is A/(x + 1) + B/(x + 1)².

Why both? A single A/(x + 1)² has a plain number on top, but (2x + 3) over (x + 1)² has x in the top. The A/(x + 1) part supplies the x. In general a factor (x + a)ⁿ on the bottom gives n parts: one for each power from 1 to n.

Worked example

Split (2x + 3)/(x + 1)² into partial fractions.

The bottom is a repeated factor, so the form is A/(x + 1) + B/(x + 1)². Both parts are needed.
Multiply both sides by (x + 1)²: 2x + 3 = A(x + 1) + B.
Put x = −1. The bracket becomes 0, so 2(−1) + 3 = B. Then B = 1.
Equate the coefficients of x: 2 = A. So A = 2.
So (2x + 3)/(x + 1)² = 2/(x + 1) + 1/(x + 1)².
Check with a number. At x = 0 the left side is 3/1 = 3. The right side is 2/1 + 1/1 = 3.

Answer: 2/(x + 1) + 1/(x + 1)²

Both sides of (2x + 3)/(x + 1)² = 2/(x + 1) + 1/(x + 1)² agree at x = 0, 1 and 2.

A bottom with both kinds of factor, such as (x + 2)(x − 1)², needs a letter for each part: A/(x + 2) + B/(x − 1) + C/(x − 1)². The cover up finds A and C. For B, equate a coefficient or put in a convenient value of x.

The cover up for (2x² + x + 3)/((x + 2)(x − 1)²). The shortcut gives A and C. It does not give B.

Worked example

Split (2x² + x + 3)/((x + 2)(x − 1)²) into partial fractions.

The form is A/(x + 2) + B/(x − 1) + C/(x − 1)².
Multiply by (x + 2)(x − 1)²: 2x² + x + 3 = A(x − 1)² + B(x + 2)(x − 1) + C(x + 2).
Put x = −2. Two terms vanish: 8 − 2 + 3 = 9 = A × (−3)² = 9A. So A = 1.
Put x = 1. Two terms vanish: 2 + 1 + 3 = 6 = C × 3. So C = 2.
Equate the coefficients of x². The left side has 2. On the right, A(x − 1)² gives A and B(x + 2)(x − 1) gives B. So 2 = A + B, and B = 1.
So the result is 1/(x + 2) + 1/(x − 1) + 2/(x − 1)².
Check at x = 0. The left side is 3/(2 × 1) = 3/2. The right side is 1/2 + 1/(−1) + 2/1 = 1/2 − 1 + 2 = 3/2.

Answer: 1/(x + 2) + 1/(x − 1) + 2/(x − 1)²

One more kind of bottom. A quadratic factor that cannot be split into real brackets, such as x² + 1, has a top with an x in it. The form is (Bx + C)/(x² + 1). A linear factor such as x − 1 still gets a plain letter: A/(x − 1).

Worked example

Split (x + 5)/((x − 1)(x² + 1)) into partial fractions.

The form is A/(x − 1) + (Bx + C)/(x² + 1).
Multiply by (x − 1)(x² + 1): x + 5 ≡ A(x² + 1) + (Bx + C)(x − 1).
Put x = 1. The second term vanishes: 6 = 2A, so A = 3.
Equate the coefficients of x². The left side has 0. The right side has A + B. So B = −3.
Equate the constants. The left side has 5. The right side has A − C. So 5 = 3 − C, and C = −2.
So the result is 3/(x − 1) + (−3x − 2)/(x² + 1), which is 3/(x − 1) − (3x + 2)/(x² + 1).
Check at x = 0. The left side is 5/((−1)(1)) = −5. The right side is 3/(−1) − 2/1 = −5.

Answer: 3/(x − 1) − (3x + 2)/(x² + 1)

Now a top that is as big as the bottom. The fraction (x² + 6x + 7)/((x + 1)(x + 2)) has a bottom of x² + 3x + 2, the same degree as the top. Divide first, as in chapter 3. The quotient is 1, and a smaller top is left over.

x² + 6x + 7 divided by x² + 3x + 2. The quotient is 1 and the remainder is 3x + 5.

Worked example

Write (x² + 6x + 7)/((x + 1)(x + 2)) as a whole number plus partial fractions.

The top has degree 2, and so does the bottom. The top is not lower, so divide first.
Expand the bottom: (x + 1)(x + 2) = x² + 3x + 2.
Divide x² + 6x + 7 by x² + 3x + 2. The quotient is 1 and the remainder is 3x + 5 (see the picture).
So the fraction equals 1 + (3x + 5)/((x + 1)(x + 2)).
Split the new fraction, as in the first example: 2/(x + 1) + 1/(x + 2).
So the fraction equals 1 + 2/(x + 1) + 1/(x + 2).
Check at x = 0. The left side is 7/2. The right side is 1 + 2/1 + 1/2 = 7/2.

Answer: 1 + 2/(x + 1) + 1/(x + 2)

Try it

Test a split with numbers

You need: paper, a pencil.

Take the split (3x + 5)/((x + 1)(x + 2)) = 2/(x + 1) + 1/(x + 2).
Put x = 0, then x = 1, then x = 2 into the left side. Write each answer as a fraction.
Put the same three values into the right side and add the two fractions.
Compare the two columns.

What you will see: The two columns agree for every value. If a split is wrong, even one value will usually show it.

If you cannot try it, this is what happens.

A wrong idea many people have

1/((x + 1)(x + 2)) = 1/(x + 1) + 1/(x + 2).

Why it feels right: It looks like splitting a product into two fractions, one for each bracket, as with numbers.

The test that breaks it: Put x = 0. The left side is 1/(1 × 2) = 1/2. The right side is 1/1 + 1/2 = 3/2. They are not equal.

What is true: The tops must be found, not guessed: 1/((x + 1)(x + 2)) = 1/(x + 1) − 1/(x + 2). Always test one value of x.

A wrong idea many people have

A repeated factor (x + 1)² needs only the single fraction A/(x + 1)².

Why it feels right: There is one factor on the bottom, so one letter seems enough.

The test that breaks it: Try 2/(x + 1)². Its top is a plain number. But (2x + 3)/(x + 1)² has x in the top, and no choice of A in A/(x + 1)² can give it.

What is true: Write A/(x + 1) + B/(x + 1)². A factor squared on the bottom needs a part for the first power and a part for the second power.

A wrong idea many people have

For (2x² + x + 3)/((x + 2)(x − 1)²), the cover up also gives B in B/(x − 1).

Why it feels right: It gave A and C so smoothly that it seems to work for every letter.

The test that breaks it: Cover (x − 1) and put x = 1. The other brackets still hold (x − 1), which is 0 there. We would divide by zero.

What is true: The cover up finds the letter for a factor that appears once, and the letter for the highest power of a repeated factor. Find the middle letter another way, such as equating the x² coefficient.

Key words

partial fractions: the plainer fractions that add up to one fraction with a product on the bottom.
cover up: a shortcut: cover one bracket and put in the x that makes it 0.
repeated factor: a bracket that appears twice or more on the bottom, such as (x + 1)².

Back to your guess

The answer is C: Yes. It becomes one fraction for each bracket on the bottom, each with a plain number on top.

It splits as 2/(x + 1) + 1/(x + 2). Each part has only one bracket on the bottom and a plain number on top. Adding them back gives the original fraction. Splitting the top, as in the second choice, is true but leaves the product on the bottom, so it is no plainer.

Keep it

Copy this into your exercise book.

One part for each bracket. A squared bracket needs A/(x + a) + B/(x + a)². Find letters by equating coefficients; the cover up checks. Divide first if the top is not lower.

Draw: The fraction on the left, an equals sign, and the two plainer fractions on the right.

M5.1.6 Chapter 6

Pascal's triangle and choosing

In how many ways can I choose 2 friends from 4?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A 8, because 4 × 2.
You picked A. Read on. At the end, the chapter comes back to your guess.
B 6.
You picked B. Read on. At the end, the chapter comes back to your guess.
C 12, because 4 × 3.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Build Pascal's triangle and link it to (a + b)ⁿ
Find nCr, the number of ways to choose r things from n

Start with powers of a + b. (a + b)¹ = a + b. And (a + b)² = a² + 2ab + b². To get (a + b)³, multiply (a + b)² by (a + b).

(a + b)(a² + 2ab + b²) = a³ + 2a²b + ab² + a²b + 2ab² + b³ = a³ + 3a²b + 3ab² + b³. Look at the numbers in front: 1 and 1, then 1, 2, 1, then 1, 3, 3, 1.

Every term of (a + b)ⁿ has total power n. The power of a goes down from n to 0, and the power of b goes up from 0 to n. Only the numbers in front, the coefficients, need work.

Pascal's triangle lists these coefficients. Row 0 is the single number 1. Every row starts and ends with 1. Every other number is the sum of the two numbers just above it. We count the rows and the places in a row from 0.

Row n gives the coefficients of (a + b)ⁿ. Row 3 reads 1, 3, 3, 1, as in the expansion above.

Each entry is the sum of the two above it: 4 + 6 = 10.

Why does the rule work? Take (a + b)ⁿ and multiply by (a + b) to get (a + b)ⁿ⁺¹. (In this book ^ means "to the power", because the font has no raised r.) A term a^(n + 1 − r) b^r in the result can arise in two ways. It comes from a times the term a^(n − r) b^r, or from b times the term a^(n + 1 − r) b^(r − 1). So its coefficient is the sum of two neighbouring coefficients of row n. That is the rule.

Worked example

Use Pascal's triangle to expand (a + b)⁵.

Row 4 is 1, 4, 6, 4, 1. Build row 5 from it. Each entry is the sum of two above it: 1 + 4 = 5, then 4 + 6 = 10.
The rest follows by symmetry. The coefficients are 1, 5, 10, 10, 5, 1.
The power of a goes down from 5 to 0, and the power of b goes up from 0 to 5. Each term has total power 5.
Write the terms: a⁵, 5a⁴b, 10a³b², 10a²b³, 5ab⁴, b⁵.
Check with a = 1 and b = 1: the left side is 2⁵ = 32. The coefficients add to 1 + 5 + 10 + 10 + 5 + 1 = 32.

Answer: a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵

The triangle gets long. For a high row, or for one entry, we want a formula. For that we count ways of choosing.

Suppose Amina, Ravi, Mei and Kofi are to be paired. Choosing 2 from 4 gives 6 pairs. In general, the number of ways to choose r things from n, when the order does not matter, is written nCr and read "n choose r". It is called a binomial coefficient.

The last row shows the raised and lowered form on a line: ⁵C₂ is the same number as 5C2. Two more ways are given in words. Other books print n and r one above the other in tall brackets. Others write n small and raised before the C, and r small and lowered after it. All mean the same number.

The 6 pairs from A, B, C and D: 3 pairs contain A, and 3 do not.

This picture shows Pascal's rule for choosing. Choose 2 from 4 people. Either A is in the pair: then choose 1 more from the other 3, which is 3C1 = 3 ways. Or A is out: choose 2 from the other 3, which is 3C2 = 3 ways. So 4C2 = 3C1 + 3C2 = 6.

In general nCr = (n − 1)C(r − 1) + (n − 1)Cr. That is the same sum as the triangle makes, so the entry in row n and place r of Pascal's triangle is nCr.

Now the formula. We need the factorial: n! = n × (n − 1) × … × 2 × 1, for a whole number n. So 5! = 5 × 4 × 3 × 2 × 1 = 120. We also define 0! = 1.

Factorials grow quickly. Note that 0! = 1, by definition.

Why define 0! = 1? There is exactly one way to choose nothing from n things. The formula below gives nC0 = n!/(0! × n!), which equals 1 only if 0! = 1.

To find a formula, choose r things in order. There are n ways for the first, n − 1 for the second, and so on down to n − r + 1 for the last: n(n − 1)…(n − r + 1) ways. Each group of r is counted r! times, once for each order. So we divide by r!.

The result is nCr = n!/(r!(n − r)!), for whole numbers n and r with 0 ≤ r ≤ n. Multiplying the top n! by 1/(n − r)! cancels the tail of the factorial: nCr = n(n − 1)…(n − r + 1)/r!.

Two facts help. nCr = nC(n − r), because choosing r to take is the same as choosing n − r to leave behind. And nC0 = nCn = 1, and nC1 = n.

See it three ways

With things

Amina, Ravi, Mei and Kofi. Choose 2 of them for a pair. List every pair: AR, AM, AK, RM, RK, MK. There are 6.

As a picture

In words

Row 4, place 2 of the triangle is the number of ways to choose 2 from 4.

In symbols

4C2 = 4!/(2! × 2!) = 24/(2 × 2) = 6, and also 4C2 = (4 × 3)/(2 × 1) = 6.

Worked example

Work out 7C3. Then write down 7C4 using symmetry.

Use nCr = n(n − 1)…(n − r + 1)/r!. For 7C3 there are 3 numbers on top, starting at 7.
7C3 = (7 × 6 × 5)/(3 × 2 × 1).
The top is 7 × 6 × 5 = 210. The bottom is 3 × 2 × 1 = 6.
210 ÷ 6 = 35. So 7C3 = 35.
Symmetry: 7C4 = 7C(7 − 4) = 7C3, because choosing 4 to take is choosing 3 to leave.
So 7C4 = 35. Check with the formula: 7!/(4! × 3!) = 5040/(24 × 6) = 5040/144 = 35.

Answer: 7C3 = 35 and 7C4 = 35.

Worked example

A club has 5 men and 3 women. A committee of 3 is chosen. How many committees are possible? How many have no women? How many have at least one woman?

Any 3 of the 8 people may be chosen, and the order does not matter. The number of committees is 8C3 = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56.
No women means all 3 come from the 5 men: 5C3 = (5 × 4 × 3)/(3 × 2 × 1) = 60/6 = 10.
"At least one woman" is every committee except the ones with no women: 56 − 10 = 46.
Check by adding the cases. One woman: 3C1 × 5C2 = 3 × 10 = 30. Two women: 3C2 × 5C1 = 3 × 5 = 15. Three women: 3C3 × 5C0 = 1. And 30 + 15 + 1 = 46.

Answer: 56 committees, 10 with no women, and 46 with at least one woman.

Worked example

(a) Show that nC2 = n(n − 1)/2. (b) Hence find n when nC2 = 28.

nC2 = n!/(2! × (n − 2)!).
Write n! = n × (n − 1) × (n − 2)!. The factor (n − 2)! cancels with the one on the bottom.
So nC2 = n(n − 1)/2!. And 2! = 2, so nC2 = n(n − 1)/2. This proves (a).
For (b), solve n(n − 1)/2 = 28. Multiply both sides by 2 and expand the bracket: n² − n = 56.
Rearrange to n² − n − 56 = 0 and factorise: (n − 8)(n + 7) = 0. So n = 8 or n = −7.
But n is a number of things, so n is a whole number from 2 up. Reject −7. So n = 8.
Check: 8C2 = (8 × 7)/2 = 28.

Answer: n = 8

The term of (a + b)⁴ with b to the power r has coefficient 4Cr. Add the five terms to get (a + b)⁴.

This links the two halves of the chapter. In (a + b)ⁿ, each term comes from choosing, in each of the n brackets, either a or b. A term with r letters b comes from choosing which r of the n brackets give b. There are nCr ways to choose them. So the coefficient is nCr.

This gives the binomial theorem for a positive whole number n: (a + b)ⁿ = aⁿ + nC1 aⁿ⁻¹b + nC2 aⁿ⁻²b² + … + bⁿ. The next chapter uses it.

Try it

Build the triangle yourself

You need: paper, a pencil.

Write a single 1 at the top. This is row 0.
Write row 1 under it: 1 and 1.
For each new row, write a 1 at each end. Between them, write the sum of the two numbers above.
Build down to row 5. Ring the entry in row 5, place 2, and check that it equals the sum of the two entries above it.

What you will see: The ringed entry is 10, and it is 4 + 6. Every entry of the triangle is a number of ways to choose.

If you cannot try it, this is what happens.

A wrong idea many people have

(a + b)² = a² + b².

Why it feels right: The square of a product splits, as in (ab)² = a²b², so the square of a sum seems to split too.

The test that breaks it: Put a = 1 and b = 1. Then (a + b)² = 4, but a² + b² = 2.

What is true: There is a middle term: (a + b)² = a² + 2ab + b². In general (a + b)ⁿ has n + 1 terms, with the coefficients from row n of Pascal's triangle.

A wrong idea many people have

0! = 0.

Why it feels right: The factorial multiplies numbers down to 1, and a product that reaches 0 is 0.

The test that breaks it: If 0! were 0, the formula for nC0 would divide by 0. But there is exactly one way to choose nothing, so nC0 = 1.

What is true: 0! = 1. It is defined this way so that the formula n!/(r!(n − r)!) works for r = 0 and r = n.

A wrong idea many people have

Choosing 2 from 4 can be done in 4 × 3 = 12 ways.

Why it feels right: There are 4 choices for the first person and 3 for the second, so multiplying looks right.

The test that breaks it: List the pairs from A, B, C and D. There are only 6: AB, AC, AD, BC, BD, CD. The product 12 counts AB and BA as different.

What is true: When the order does not matter, divide by the number of orders, r!. So 4C2 = (4 × 3)/2! = 6.

A wrong idea many people have

The numbers 1, 3, 3, 1 are row 4, so they give the coefficients of (a + b)⁴.

Why it feels right: There are four numbers, and 4 is the number of numbers.

The test that breaks it: Check (a + b)³ = a³ + 3a²b + 3ab² + b³. The coefficients are 1, 3, 3, 1. And (a + b)⁴ has five terms.

What is true: Count the rows from 0. Row n has n + 1 numbers and gives the coefficients of (a + b)ⁿ. So 1, 3, 3, 1 is row 3.

Key words

Pascal's triangle: a triangle of numbers where each entry is the sum of the two above it. Row n gives the coefficients of (a + b)ⁿ.
binomial coefficient: the number nCr, the coefficient of a^(n − r) b^r in (a + b)ⁿ.
factorial: n! = n × (n − 1) × … × 1, with 0! = 1.

Back to your guess

The answer is B: 6.

Call the friends A, B, C and D. The pairs are AB, AC, AD, BC, BD and CD: 6 pairs. The sum 4 × 3 = 12 counts every pair twice, once in each order, and "A then B" is the same pair as "B then A". The number of ways to choose r things from n is written nCr, and 4C2 = 6.

Keep it

Copy this into your exercise book.

Pascal's triangle: each entry is the sum of the two above; row n gives (a + b)ⁿ. nCr = n!/(r!(n − r)!) counts choices; 0! = 1. nCr = nC(n − r).

Draw: The first six rows of Pascal's triangle, with one entry shown as the sum of the two above it.

M5.1.7 Chapter 7

The binomial expansion

Can I expand (1 + x)⁻¹ the way I expand (1 + x)²?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

A No. A power that is not a positive whole number cannot be expanded.
You picked A. Read on. At the end, the chapter comes back to your guess.
B Yes, and the series ends after a few terms, as it does for (1 + x)².
You picked B. Read on. At the end, the chapter comes back to your guess.
C Yes, but the series never ends, and it works only when x is small enough.
You picked C. Read on. At the end, the chapter comes back to your guess.

By the end you can

Expand (2 + x)⁵ with the binomial theorem
Find one term, like the x³ term, without the full expansion
Expand (1 + x)⁻¹ for small x and say when it is valid

Chapter 6 showed that the coefficient of a^(n − r) b^r in (a + b)ⁿ is nCr. (The sign ^ means "to the power". The book font has no raised r.) Written out, this is the binomial theorem.

(a + b)ⁿ = aⁿ + nC1 aⁿ⁻¹ b + nC2 aⁿ⁻² b² + nC3 aⁿ⁻³ b³ + … + bⁿ. It holds for every positive whole number n. The term with b to the power r is nCr a^(n − r) b^r.

Take a = 2 and b = x. Each term is nCr × 2 to the power (n − r) × x to the power r. The powers of 2 must be worked out. They are the usual place for a slip.

Row 5 holds the numbers 5C0 to 5C5. They are the coefficients before the powers of 2 are put in.

The five columns of work for (2 + x)⁵. The term is 5Cr times the power of 2 times x to the power r.

Worked example

Expand (2 + x)⁵ in ascending powers of x. (Ascending means the lowest power of x first.)

Here a = 2, b = x and n = 5. The coefficients 5C0 to 5C5 are 1, 5, 10, 10, 5, 1 (row 5 of Pascal's triangle).
The power of 2 falls from 5 to 0: 2⁵ = 32, 2⁴ = 16, 2³ = 8, 2² = 4, 2¹ = 2, 2⁰ = 1.
Multiply each coefficient by its power of 2: 1 × 32 = 32, 5 × 16 = 80, 10 × 8 = 80, 10 × 4 = 40, 5 × 2 = 10, 1 × 1 = 1.
Attach x to the power r: 32 + 80x + 80x² + 40x³ + 10x⁴ + x⁵.
Check with x = 1. The left side is 3⁵ = 243. The right side is 32 + 80 + 80 + 40 + 10 + 1 = 243.

Answer: 32 + 80x + 80x² + 40x³ + 10x⁴ + x⁵

Often we want only one term. The general term, the one with b to the power r, is the (r + 1)th term. It is written T(r + 1) = nCr a^(n − r) b^r. Put in the values of n, r, a and b. There is no need to write the whole expansion.

Take care when b has a number in it. The whole of b is raised to the power r. So (2x)³ = 2³ × x³ = 8x³, and (−3x)² = 9x².

The three parts of the x³ term of (3 + 2x)⁵, and their product.

Worked example

Find the coefficient of x³ in the expansion of (3 + 2x)⁵.

Here a = 3, b = 2x and n = 5. The term with x³ has r = 3.
The general term is 5C3 × 3² × (2x)³, because the power of a is n − r = 2.
Work out the parts: 5C3 = 10, 3² = 9, and (2x)³ = 8x³.
Multiply: 10 × 9 × 8 = 720.
So the x³ term is 720x³, and the coefficient is 720.

Answer: The coefficient of x³ is 720.

Worked example

Find the coefficient of x² in the expansion of (1 + 2x)(1 − x)⁶.

First expand (1 − x)⁶ as far as x². Here a = 1 and b = −x, so the terms are 6C0, 6C1(−x), 6C2(−x)².
6C0 = 1. 6C1 × (−x) = −6x. 6C2 × (−x)² = 15x², because (−x)² = x².
So (1 − x)⁶ = 1 − 6x + 15x² + …
Now multiply by (1 + 2x). The x² terms come from two products: 1 × 15x², and 2x × (−6x).
These are 15x² and −12x². Their sum is 15 − 12 = 3.
So the coefficient of x² is 3.

Answer: The coefficient of x² is 3.

Now the power can be any rational number, not only a positive whole number. A rational number is a fraction of two whole numbers, with the bottom not zero. This includes −1, −2 and 1/2.

The coefficients nC1, nC2, nC3 are n, n(n − 1)/2!, n(n − 1)(n − 2)/3!. These products make sense for every n, not only whole numbers. This gives the general result.

For any rational n, (1 + x)ⁿ = 1 + nx + [n(n − 1)/2!] x² + [n(n − 1)(n − 2)/3!] x³ + …, and it is valid for |x| < 1, which means −1 < x < 1.

There are two big differences from a positive whole number n. First, the series never stops: for n = −1 no factor n − r is ever 0. Second, it works only for |x| < 1. For a positive whole n, the factor (n − n) = 0 stops the series after n + 1 terms, and then it is true for every x.

Take n = −1. The coefficients are 1, then n = −1, then (−1)(−2)/2! = 1, then (−1)(−2)(−3)/3! = −1. So (1 + x)⁻¹ = 1 − x + x² − x³ + …

Why |x| < 1? The terms 1, −x, x², −x³ are a geometric series with first term 1 and common ratio −x. A geometric series adds up to 1/(1 − (−x)) = 1/(1 + x) only when the ratio is between −1 and 1, that is, |x| < 1.

Four terms of the series for 1/(1 + x). The sum is close when x is small, rough at x = 0.5, and far wrong at x = 2.

f(x) = 1/(1 + x) and g(x) = 1 − x + x² − x³. Near x = 0 the two agree. Towards x = −1 the series falls far short.

Worked example

Expand (1 + x)⁻¹ in ascending powers of x as far as x³. State the values of x for which the expansion is valid, and test it at x = 0.1.

Use (1 + x)ⁿ = 1 + nx + n(n − 1)/2! x² + n(n − 1)(n − 2)/3! x³ + … with n = −1.
The x term is nx = −x.
The x² coefficient is (−1)(−2)/2 = 1.
The x³ coefficient is (−1)(−2)(−3)/6 = −1.
So (1 + x)⁻¹ = 1 − x + x² − x³ + …, valid for |x| < 1.
Test at x = 0.1: the left side is 1/1.1 = 0.90909… The right side is 1 − 0.1 + 0.01 − 0.001 = 0.909, close to the left.

Answer: 1 − x + x² − x³ + …, valid for |x| < 1.

To expand (a + bx)ⁿ for a negative or fractional n, first take out the number a: (a + bx)ⁿ = aⁿ(1 + (b/a)x)ⁿ. Then expand (1 + (b/a)x)ⁿ. This needs a > 0 when n is not a whole number, so that aⁿ makes sense. The expansion is valid when |(b/a)x| < 1, which means |x| < |a/b|.

For (1 + 3x)⁻², put 3x in place of x. The series is valid for |3x| < 1, which means |x| < 1/3. The limit is 1/3, not 1. Always divide: the limit for |bx| < 1 is |x| < 1/|b|.

Worked example

Expand (1 + 3x)⁻² in ascending powers of x up to x³. State the range of x for which it is valid. Hence find the coefficient of x² in the expansion of (3 − x)(1 + 3x)⁻².

Put 3x in place of x and n = −2 in (1 + x)ⁿ. The coefficients are n = −2, then (−2)(−3)/2 = 3, then (−2)(−3)(−4)/6 = −4.
The terms are 1, then −2 × 3x = −6x, then 3 × (3x)² = 3 × 9x² = 27x², then −4 × (3x)³ = −4 × 27x³ = −108x³.
So (1 + 3x)⁻² = 1 − 6x + 27x² − 108x³ + …
The expansion is valid for |3x| < 1, which means |x| < 1/3.
For the product (3 − x)(1 + 3x)⁻², the x² terms are 3 × 27x² and (−x) × (−6x).
These are 81x² and 6x². So the coefficient of x² is 81 + 6 = 87.

Answer: 1 − 6x + 27x² − 108x³, valid for |x| < 1/3, and the coefficient is 87.

A case worth knowing is n = 1/2, the square root. Its coefficients are 1, then n = 1/2, then (1/2)(−1/2)/2! = −1/8, then (1/2)(−1/2)(−3/2)/3! = 1/16. So √(1 + x) = 1 + x/2 − x²/8 + x³/16 − …, valid for |x| < 1. For example √1.02 is about 1 + 0.01 − 0.00005 = 1.00995.

Worked example

Expand (1 − 2x)⁻¹ in ascending powers of x up to x³. State when it is valid.

We know (1 + y)⁻¹ = 1 − y + y² − y³ + …, valid for |y| < 1.
Put y = −2x. Then −y = 2x, y² = 4x² and −y³ = 8x³.
So (1 − 2x)⁻¹ = 1 + 2x + 4x² + 8x³ + …
The condition |y| < 1 becomes |−2x| < 1, which is |2x| < 1. So |x| < 1/2.
Check with x = 0.1. The series gives 1 + 0.2 + 0.04 + 0.008 = 1.248. The left side is 1/0.8 = 1.25, which is close.

Answer: 1 + 2x + 4x² + 8x³ + …, valid for |x| < 1/2.

Worked example

Expand √(4 − x) in ascending powers of x up to x². State when it is valid. Use x = 0.04 to estimate √3.96.

Take out the 4: √(4 − x) = √4 × √(1 − x/4) = 2(1 − x/4)^(1/2).
Use (1 + y)ⁿ with n = 1/2 and y = −x/4: 1 + ny + n(n − 1)/2! y² + …
The y term is (1/2)(−x/4) = −x/8.
The y² coefficient is (1/2)(−1/2)/2 = −1/8. And y² = x²/16, so the term is −x²/128.
So (1 − x/4)^(1/2) = 1 − x/8 − x²/128 + … Multiply by 2: √(4 − x) = 2 − x/4 − x²/64 + …
It is valid for |−x/4| < 1, which means |x| < 4.
Put x = 0.04. Then √3.96 is about 2 − 0.04/4 − 0.0016/64 = 2 − 0.01 − 0.000025 = 1.989975.

Answer: 2 − 1/4x − 1/64x², valid for |x| < 4, and √3.96 is about 1.989975.

Another favourite is the term independent of x, the one with no x in it. In a general term, collect the power of x, set it equal to 0, and solve for r. If r is not a whole number from 0 to n, there is no such term.

Worked example

Find the term independent of x in the expansion of (x + 2/x)⁶.

Here a = x, b = 2/x and n = 6. The general term is T(r + 1) = 6Cr × x^(6 − r) × (2/x)^r.
Write (2/x)^r = 2^r × x^(−r). So the term is 6Cr × 2^r × x^(6 − r − r) = 6Cr × 2^r × x^(6 − 2r).
The term has no x when the power is 0: 6 − 2r = 0, so r = 3.
Put r = 3: 6C3 = 20 and 2³ = 8.
So the term is 20 × 8 = 160, and that is the term independent of x.

Answer: The term independent of x is 160.

Try it

Test an expansion with numbers

You need: paper, a pencil, a calculator if you have one.

Take the result (2 + x)⁵ = 32 + 80x + 80x² + 40x³ + 10x⁴ + x⁵.
Put x = 1, then x = 2, then x = −1 into the left side. Write each answer.
Put the same values into the right side and add the terms.
Compare. A wrong coefficient will almost always change one of the answers.

What you will see: The two sides agree for all three values. Putting in x = 1 is the quickest check, because it just adds the coefficients.

If you cannot try it, this is what happens.

A wrong idea many people have

(2 + x)⁵ = 2 + 5x + 10x² + 10x³ + 5x⁴ + x⁵, using the row of the triangle only.

Why it feels right: The coefficients of (1 + x)⁵ are 1, 5, 10, 10, 5, 1, and the 2 seems to be just one more number.

The test that breaks it: Put x = 1. The left side is 3⁵ = 243. This right side adds to 2 + 5 + 10 + 10 + 5 + 1 = 33, which is not 243.

What is true: Every term needs the power of 2: 2⁵, 2⁴, 2³, and so on. The coefficient of x is 5 × 2⁴ = 80, not 5.

A wrong idea many people have

In (3 + 2x)⁵ the x³ term uses 2x³, because the 2 is just a number in front.

Why it feels right: The 2 is next to the x, so only x seems to get the power 3.

The test that breaks it: (2x)³ = 2x × 2x × 2x = 8x³. The whole of 2x is cubed, so 2 is cubed too.

What is true: Raise the whole of b to the power r: (2x)³ = 8x³. Then the coefficient is 10 × 3² × 8 = 720, not 10 × 9 × 2 = 180.

A wrong idea many people have

The series for (1 + x)⁻¹ gives the right answer for every x.

Why it feels right: For a positive whole n, the expansion is true for every x, so it seems the same for all n.

The test that breaks it: Put x = 2. The series gives 1 − 2 + 4 − 8 + 16 − …, which grows without limit. But 1/(1 + 2) = 1/3.

What is true: For n not a positive whole number, the series is valid only for |x| < 1, and for (1 + bx)ⁿ only for |bx| < 1. State the range every time.

A wrong idea many people have

The expansion of (1 + x)⁻¹ stops after a few terms, as (1 + x)³ does.

Why it feels right: Any power of a bracket seems to have a last term, as bⁿ does.

The test that breaks it: The coefficients for n = −1 are 1, −1, 1, −1, … They are never 0. For n = 3 the factor (n − 3) is 0, which stops the series.

What is true: Only a positive whole number n gives a finite expansion. Any other n gives a series that goes on for ever, which is why its range of validity must be stated.

Key words

binomial theorem: the formula for (a + b)ⁿ as a sum of terms nCr a^(n − r) b^r, for a positive whole number n.
general term: the term with b to the power r in the expansion, which lets you find one term without the rest.
ascending powers: written with the lowest power of x first.
valid: giving the right value. The series for (1 + x)ⁿ is valid only for |x| < 1 unless n is a positive whole number.

Back to your guess

The answer is C: Yes, but the series never ends, and it works only when x is small enough.

It can be expanded: 1/(1 + x) = 1 − x + x² − x³ + …. The terms go on for ever, so the series never ends. And it gives the right value only when x lies between −1 and 1. For x = 2 the terms 1, −2, 4, −8 grow without limit, but 1/(1 + 2) is just a third.

Keep it

Copy this into your exercise book.

(a + b)ⁿ: term nCr a^(n − r) b^r. Raise all of b to r. For any other n: (1 + x)ⁿ = 1 + nx + n(n − 1)/2! x² + …, endless, valid for |x| < 1.

Draw: A line of expansion terms with nCr above each, and the number line from −1 to 1 marked as the valid range.

Fold the graph along y = x

Fold up the part below the axis

Divide x³ + 2x² − 5x − 6 by x + 1

Multiply back to check a division

Find where a cubic crosses

Test a split with numbers

Build the triangle yourself

Test an expansion with numbers

Key words, A to Z