Chapter 5 of 7

All 7 chapters
  1. 1Domain, range and inverses, readunderstood
  2. 2The modulus function, readunderstood
  3. 3Dividing polynomials, readunderstood
  4. 4The factor theorem, readunderstood
  5. 5Partial fractions, readunderstood
  6. 6Pascal's triangle and choosing, readunderstood
  7. 7The binomial expansion, readunderstood

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

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • 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.

A table with the heads cover, put x = and value. Row one: cover x + 1, put x = −1, value 2. Row two: cover x + 2, put x = −2, value 1.coverput x =valuex + 1−12x + 2−21
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.

  1. Write (3x + 5)/((x + 1)(x + 2)) = A/(x + 1) + B/(x + 2).
  2. Multiply both sides by (x + 1)(x + 2). The bottoms cancel: 3x + 5 = A(x + 2) + B(x + 1).
  3. Expand the right side: 3x + 5 = (A + B)x + (2A + B).
  4. Equate the coefficients of x: A + B = 3. Equate the constants: 2A + B = 5.
  5. Subtract the first equation from the second: A = 2. Then B = 3 − 2 = 1.
  6. So (3x + 5)/((x + 1)(x + 2)) = 2/(x + 1) + 1/(x + 2).
  7. 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.
  8. 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.

  1. The bottom is a repeated factor, so the form is A/(x + 1) + B/(x + 1)². Both parts are needed.
  2. Multiply both sides by (x + 1)²: 2x + 3 = A(x + 1) + B.
  3. Put x = −1. The bracket becomes 0, so 2(−1) + 3 = B. Then B = 1.
  4. Equate the coefficients of x: 2 = A. So A = 2.
  5. So (2x + 3)/(x + 1)² = 2/(x + 1) + 1/(x + 1)².
  6. 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

A table with the heads x, left and right. Row one: x = 0, left 3, right 3. Row two: x = 1, left 5/4, right 5/4. Row three: x = 2, left 7/9, right 7/9.xleftright03315/45/427/97/9
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.

A table with the heads cover, put x = and value. Row one: cover x + 2, put x = −2, value 1. Row two: cover (x − 1)², put x = 1, value 2.coverput x =valuex + 2−21(x − 1)²12
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.

  1. The form is A/(x + 2) + B/(x − 1) + C/(x − 1)².
  2. Multiply by (x + 2)(x − 1)²: 2x² + x + 3 = A(x − 1)² + B(x + 2)(x − 1) + C(x + 2).
  3. Put x = −2. Two terms vanish: 8 − 2 + 3 = 9 = A × (−3)² = 9A. So A = 1.
  4. Put x = 1. Two terms vanish: 2 + 1 + 3 = 6 = C × 3. So C = 2.
  5. 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.
  6. So the result is 1/(x + 2) + 1/(x − 1) + 2/(x − 1)².
  7. 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.

  1. The form is A/(x − 1) + (Bx + C)/(x² + 1).
  2. Multiply by (x − 1)(x² + 1): x + 5 ≡ A(x² + 1) + (Bx + C)(x − 1).
  3. Put x = 1. The second term vanishes: 6 = 2A, so A = 3.
  4. Equate the coefficients of x². The left side has 0. The right side has A + B. So B = −3.
  5. Equate the constants. The left side has 5. The right side has A − C. So 5 = 3 − C, and C = −2.
  6. So the result is 3/(x − 1) + (−3x − 2)/(x² + 1), which is 3/(x − 1) − (3x + 2)/(x² + 1).
  7. 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.

A long division. The divisor x² + 3x + 2 stands left of a bracket over x² + 6x + 7. The quotient above is 1. Under the dividend is the row x² + 3x + 2, and under it the remainder 3x + 5.x² + 3x + 2x²+ 6x+ 71x²+ 3x+ 23x+ 5
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.

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

  1. Take the split (3x + 5)/((x + 1)(x + 2)) = 2/(x + 1) + 1/(x + 2).
  2. Put x = 0, then x = 1, then x = 2 into the left side. Write each answer as a fraction.
  3. Put the same three values into the right side and add the two fractions.
  4. 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 table with the heads x, left and right. Row one: x = 0, left 5/2, right 5/2. Row two: x = 1, left 4/3, right 4/3. Row three: x = 2, left 11/12, right 11/12.xleftright05/25/214/34/3211/1211/12

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.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. Which is the correct form of the partial fractions of 5/((x − 1)(x + 3))?

    Tap your answer.

    1. Not this one. That is the fraction itself with a new letter. The bottom must be split, with one bracket for each part. Go to M5.1.5. Try another.

    2. Not this one. That keeps the whole bottom, so nothing is plainer. Use one fraction for each bracket. Go to M5.1.5. Try another.

    3. Not this one. Squares are used only for a repeated factor. Here each bracket appears once. Go to M5.1.5. Try another.

    4. Yes, that is right. The bottom has two different brackets, each once. Each gets its own fraction with a plain letter on top.

    See the answer

    A/(x − 1) + B/(x + 3)

    The bottom has two different brackets, each once. Each gets its own fraction with a plain letter on top.

    • If you chose A: That is the fraction itself with a new letter. The bottom must be split, with one bracket for each part. Go to M5.1.5.
    • If you chose B: That keeps the whole bottom, so nothing is plainer. Use one fraction for each bracket. Go to M5.1.5.
    • If you chose C: Squares are used only for a repeated factor. Here each bracket appears once. Go to M5.1.5.

  2. Which is the correct form of the partial fractions of 4/(x + 2)²?

    Tap your answer.

    1. Yes, that is right. A factor (x + 2)² needs one part for each power from 1 to 2: A/(x + 2) and B/(x + 2)².

    2. Not this one. That has a part for the square only. A repeated factor needs a part for the first power as well. Go to M5.1.5. Try another.

    3. Not this one. The second bracket must be the same one, not a new bracket. A repeated factor is a power of one bracket. Go to M5.1.5. Try another.

    4. Not this one. The square (x + 2)² is not (x + 2)(x + 4). The second part uses the same bracket, squared. Go to M5.1.5. Try another.

    See the answer

    A/(x + 2) + B/(x + 2)²

    A factor (x + 2)² needs one part for each power from 1 to 2: A/(x + 2) and B/(x + 2)².

    • If you chose B: That has a part for the square only. A repeated factor needs a part for the first power as well. Go to M5.1.5.
    • If you chose C: The second bracket must be the same one, not a new bracket. A repeated factor is a power of one bracket. Go to M5.1.5.
    • If you chose D: The square (x + 2)² is not (x + 2)(x + 4). The second part uses the same bracket, squared. Go to M5.1.5.

  3. The table shows the cover up for (5x + 1)/((x − 1)(x + 2)) = A/(x − 1) + B/(x + 2). What is B?

    A table with the heads cover, put x = and value. Row one: cover x − 1, put x = 1, value 2. Row two: cover x + 2, put x = −2, value left empty.coverput x =valuex − 112x + 2−2
    Tap your answer.

    1. Not this one. Check the signs: at x = −2 the top is 5(−2) + 1 = −9 and the other bracket is x − 1 = −3. Then −9 ÷ −3 is positive. Go to M5.1.5. Try another.

    2. Not this one. You took the size of the top, 9, and did not divide by the other bracket. At x = −2 that bracket is x − 1 = −3. Go to M5.1.5. Try another.

    3. Yes, that is right. Cover x + 2 and put x = −2. The top is 5(−2) + 1 = −9. The other bracket is x − 1 = −2 − 1 = −3. So B = −9 ÷ (−3) = 3.

    4. Not this one. That is only the top. Divide by the other bracket x − 1, which is −3 when x = −2. Go to M5.1.5. Try another.

    See the answer

    3

    Cover x + 2 and put x = −2. The top is 5(−2) + 1 = −9. The other bracket is x − 1 = −2 − 1 = −3. So B = −9 ÷ (−3) = 3.

    • If you chose A: Check the signs: at x = −2 the top is 5(−2) + 1 = −9 and the other bracket is x − 1 = −3. Then −9 ÷ −3 is positive. Go to M5.1.5.
    • If you chose B: You took the size of the top, 9, and did not divide by the other bracket. At x = −2 that bracket is x − 1 = −3. Go to M5.1.5.
    • If you chose D: That is only the top. Divide by the other bracket x − 1, which is −3 when x = −2. Go to M5.1.5.

  4. How many letters are needed to split (2x² + x + 3)/((x + 2)(x − 1)²) into partial fractions?

    Tap your answer.

    1. Not this one. You counted one letter for each bracket. The squared bracket needs a letter for each of its two powers. Go to M5.1.5. Try another.

    2. Yes, that is right. The bottom has the bracket x + 2 once, and the bracket x − 1 twice. That gives 1 + 2 = 3 parts, so 3 letters: A/(x + 2) + B/(x − 1) + C/(x − 1)².

    3. Not this one. You counted the powers 2 and 2 together. The first bracket appears once, so it has only one part. Go to M5.1.5. Try another.

    4. Not this one. One letter would give one fraction, which cannot change the bottom. Count the parts: one for each power of each bracket. Go to M5.1.5. Try another.

    See the answer

    3

    The bottom has the bracket x + 2 once, and the bracket x − 1 twice. That gives 1 + 2 = 3 parts, so 3 letters: A/(x + 2) + B/(x − 1) + C/(x − 1)².

    • If you chose A: You counted one letter for each bracket. The squared bracket needs a letter for each of its two powers. Go to M5.1.5.
    • If you chose C: You counted the powers 2 and 2 together. The first bracket appears once, so it has only one part. Go to M5.1.5.
    • If you chose D: One letter would give one fraction, which cannot change the bottom. Count the parts: one for each power of each bracket. Go to M5.1.5.

  5. Tariq splits 1/((x + 1)(x + 2)) as 1/(x + 1) + 1/(x + 2). What is wrong?

    Tap your answer.

    1. Yes, that is right. Over the common bottom, 1/(x + 1) + 1/(x + 2) = ((x + 2) + (x + 1))/((x + 1)(x + 2)) = (2x + 3)/((x + 1)(x + 2)). The top should be 1, so the tops are wrong. The correct split is 1/(x + 1) − 1/(x + 2).

    2. Not this one. Try x = 0. The left side is 1/2, but his right side is 1 + 1/2. A split must give the same value for every x. Go to M5.1.5. Try another.

    3. Not this one. Bottoms are not added. In partial fractions each bracket keeps its own fraction, and the tops are found by solving. Go to M5.1.5. Try another.

    See the answer

    Adding his two fractions gives a top of 2x + 3, not 1.

    Over the common bottom, 1/(x + 1) + 1/(x + 2) = ((x + 2) + (x + 1))/((x + 1)(x + 2)) = (2x + 3)/((x + 1)(x + 2)). The top should be 1, so the tops are wrong. The correct split is 1/(x + 1) − 1/(x + 2).

    • If you chose B: Try x = 0. The left side is 1/2, but his right side is 1 + 1/2. A split must give the same value for every x. Go to M5.1.5.
    • If you chose C: Bottoms are not added. In partial fractions each bracket keeps its own fraction, and the tops are found by solving. Go to M5.1.5.

  6. Split (7x + 11)/((x + 1)(x + 3)) into A/(x + 1) + B/(x + 3). Find A and B. Give A first, then B.

    See the answer

    A = 2, B = 5

    Cover x + 1 and put x = −1: (7(−1) + 11)/(−1 + 3) = 4/2 = 2. Cover x + 3 and put x = −3: (7(−3) + 11)/(−3 + 1) = (−10)/(−2) = 5. Check at x = 0: 11/3 on the left, and 2/1 + 5/3 = 11/3 on the right.

  7. Split (2x + 5)/(x + 1)² into A/(x + 1) + B/(x + 1)². Find A and B. Give A first, then B.

    See the answer

    A = 2, B = 3

    Multiply by (x + 1)²: 2x + 5 = A(x + 1) + B. Put x = −1: 2(−1) + 5 = B, so B = 3. Equate the coefficients of x: A = 2. Check at x = 0: 5 on the left, and 2/1 + 3/1 = 5 on the right.

  8. Write (x² + 8x + 13)/((x + 2)(x + 3)) in the form 1 + A/(x + 2) + B/(x + 3). Find A and B. Give A first, then B.

    See the answer

    A = 1, B = 2

    The bottom is x² + 5x + 6. Take it from the top: x² + 8x + 13 − (x² + 5x + 6) = 3x + 7. So the fraction is 1 + (3x + 7)/((x + 2)(x + 3)). Cover x + 2 and put x = −2: (3(−2) + 7)/(−2 + 3) = 1. Cover x + 3 and put x = −3: (3(−3) + 7)/(−3 + 2) = (−2)/(−1) = 2.

  9. (a) Show that 1/(x(x + 1)) = 1/x − 1/(x + 1). (b) Use this to find 1/(1 × 2) + 1/(2 × 3) + 1/(3 × 4) as a single fraction.

    See the answer

    3/4

    (a) Put 1/x − 1/(x + 1) over the common bottom: ((x + 1) − x)/(x(x + 1)) = 1/(x(x + 1)). (b) Put x = 1, 2, 3: the sum is (1 − 1/2) + (1/2 − 1/3) + (1/3 − 1/4). Each middle term cancels, which leaves 1 − 1/4 = 3/4.

  10. From an earlier chapter

    Which of these is a factor of x³ − 2x² + x − 2?

    Tap your answer.

    1. Not this one. Test f(−2) = −8 − 8 − 2 − 2 = −20, which is not 0. The sign flips: x + 2 is tested with −2. Go to M5.1.4. Try another.

    2. Not this one. Test f(1) = 1 − 2 + 1 − 2 = −2, which is not 0, so x − 1 is not a factor. Go to M5.1.4. Try another.

    3. Not this one. Test f(−1) = −1 − 2 − 1 − 2 = −6, which is not 0, so x + 1 is not a factor. Go to M5.1.4. Try another.

    4. Yes, that is right. Test f(2) = 8 − 8 + 2 − 2 = 0. So x − 2 is a factor, by the factor theorem. The other three values give −20, −2 and −6.

    See the answer

    x − 2

    Test f(2) = 8 − 8 + 2 − 2 = 0. So x − 2 is a factor, by the factor theorem. The other three values give −20, −2 and −6.

    • If you chose A: Test f(−2) = −8 − 8 − 2 − 2 = −20, which is not 0. The sign flips: x + 2 is tested with −2. Go to M5.1.4.
    • If you chose B: Test f(1) = 1 − 2 + 1 − 2 = −2, which is not 0, so x − 1 is not a factor. Go to M5.1.4.
    • If you chose C: Test f(−1) = −1 − 2 − 1 − 2 = −6, which is not 0, so x + 1 is not a factor. Go to M5.1.4.

  11. Split (3x² + 2x − 1)/((x − 1)(x² + 1)) into A/(x − 1) + (Bx + C)/(x² + 1). Find A, B and C. Give them in that order.

    See the answer

    A = 2, B = 1, C = 3

    Multiply by (x − 1)(x² + 1): 3x² + 2x − 1 ≡ A(x² + 1) + (Bx + C)(x − 1). Put x = 1: 4 = 2A, so A = 2. Equate x²: 3 = A + B, so B = 1. Equate the constants: −1 = A − C, so C = 3. Check at x = 0: −1/(−1 × 1) = 1 on the left, and 2/(−1) + 3/1 = 1 on the right.

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.

Reading with a child? A short page for the adult: what this chapter teaches and three questions to ask.

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