Chapter 4 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

M5.1.4 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

  • 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

  • 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.

A table with the heads a and f(a). The rows are a = 1, f(a) = 6; a = −1, f(a) = 12; a = 2, f(a) = 0.af(a)16−11220
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.

A long division. The divisor x − 2 stands left of a bracket over x³ − 3x² − 4x + 12. The quotient above is x² − x − 6. The last line is the remainder 0.x − 2x³− 3x²− 4x+ 12x²x³− 2x²−x²− 4x− x−x²+ 2x−6x+ 12− 6−6x+ 120
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.

  1. Work out f(2) = 2³ − 3 × 2² − 4 × 2 + 12 = 8 − 12 − 8 + 12 = 0.
  2. Since f(2) = 0, the factor theorem says x − 2 is a factor.
  3. Divide f(x) by x − 2. The quotient is x² − x − 6 (see the picture) and the remainder is 0.
  4. Factorise the quadratic: find two numbers with product −6 and sum −1. They are −3 and 2.
  5. So x² − x − 6 = (x − 3)(x + 2).
  6. So f(x) = (x − 2)(x − 3)(x + 2). Written with the roots in order: (x + 2)(x − 2)(x − 3).
  7. 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.

A graph with x from −3 to 4 and y from −2 to 16. A curve named y = f(x) rises from the bottom left, crosses the x axis at −2, climbs past the point (0, 12), then falls to cross the x axis at 2 and again at 3, and then rises steeply. Dots mark the three crossings, at −2, 2 and 3 on the x axis, and the point (0, 12) named on the y axis.(0, 12)y = f(x)−3−2−11234−22468101214160xy
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.

  1. 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.
  2. Put x = 0: y = 0³ − 3 × 0² − 4 × 0 + 12 = 12. The curve meets the y axis at (0, 12).
  3. 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.
  4. Between the roots the sign changes. The curve is above the axis between −2 and 2, and below the axis between 2 and 3.
  5. 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.

A graph with x from −3 to 3 and y from −4 to 6. A curve named y = f(x) rises from the bottom left, crosses the x axis at −2, rises to a hump, comes down to touch the x axis at 1 without crossing, and then rises again. Dots mark the point (−2, 0) where it crosses, the point (1, 0) where it touches, and the point (0, 2) on the y axis.(−2, 0)(1, 0)(0, 2)y = f(x)−3−2−1123−4−22460xy
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.

  1. f(2) = 2 × 8 + 4 − 26 + 6 = 16 + 4 − 26 + 6 = 0.
  2. Since f(2) = 0, the factor theorem says x − 2 is a factor.
  3. Divide f(x) by x − 2. The quotient is 2x² + 5x − 3 (see the picture).
  4. Factorise 2x² + 5x − 3. Find two numbers with product 2 × (−3) = −6 and sum 5. They are 6 and −1.
  5. Split the middle term: 2x² + 6x − x − 3 = 2x(x + 3) − (x + 3) = (2x − 1)(x + 3).
  6. So f(x) = (x − 2)(2x − 1)(x + 3).
  7. 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.

A long division. The divisor x − 2 stands left of a bracket over 2x³ + x² − 13x + 6. The quotient above is 2x² + 5x − 3. The last line is the remainder 0.x − 22x³+ x²− 13x+ 62x²2x³− 4x²5x²− 13x+ 5x5x²− 10x−3x+ 6− 3−3x+ 60
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.

A graph with x from −2 to 4 and y from −12 to 12. A curve named y = f(x) rises from the bottom left and crosses the x axis once, at 3, then keeps rising. It meets the y axis at the point (0, −6), and a dot marks the point (3, 0).(3, 0)(0, −6)y = f(x)−2−11234−12−8−448120xy
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.

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

  1. Make a table of f(x) = x³ − 3x² − 4x + 12 for x from −3 to 4.
  2. Look down the f(x) column. Mark every x where f(x) is 0.
  3. Plot the points and join them with a smooth curve.
  4. 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 table with the heads x and f(x). The rows are x = −3, f(x) = −30; −2, 0; −1, 12; 0, 12; 1, 6; 2, 0; 3, 0; 4, 12.xf(x)−3−30−20−112012162030412

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.

Check yourself

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

  1. Which of these is a factor of f(x) = x³ + x² − 10x + 8?

    Tap your answer.

    1. Not this one. That needs f(−1) = 0, but f(−1) = −1 + 1 + 10 + 8 = 18. The sign flips: x + 1 is tested with −1. Go to M5.1.4. Try another.

    2. Not this one. That needs f(−2) = 0, but f(−2) = −8 + 4 + 20 + 8 = 24, which is not 0. Go to M5.1.4. Try another.

    3. Not this one. That needs f(3) = 0, but f(3) = 27 + 9 − 30 + 8 = 14. The number 3 does not divide the constant 8. Go to M5.1.4. Try another.

    4. Yes, that is right. Test f(1) = 1 + 1 − 10 + 8 = 0. Since f(1) = 0, x − 1 is a factor.

    See the answer

    x − 1

    Test f(1) = 1 + 1 − 10 + 8 = 0. Since f(1) = 0, x − 1 is a factor.

    • If you chose A: That needs f(−1) = 0, but f(−1) = −1 + 1 + 10 + 8 = 18. The sign flips: x + 1 is tested with −1. Go to M5.1.4.
    • If you chose B: That needs f(−2) = 0, but f(−2) = −8 + 4 + 20 + 8 = 24, which is not 0. Go to M5.1.4.
    • If you chose C: That needs f(3) = 0, but f(3) = 27 + 9 − 30 + 8 = 14. The number 3 does not divide the constant 8. Go to M5.1.4.

  2. x + 2 is a factor of x³ + 3x² + kx + 2. What is k?

    Tap your answer.

    1. Not this one. You may have written 6 + 2k = 0. The term is kx, and x = −2 gives −2k, so the equation is 6 − 2k = 0. Go to M5.1.4. Try another.

    2. Yes, that is right. Since x + 2 is a factor, f(−2) = 0. f(−2) = −8 + 12 − 2k + 2 = 6 − 2k. So 6 − 2k = 0, and k = 3.

    3. Not this one. You put in x = 2. The factor is x + 2, which is x − (−2), so use x = −2. Go to M5.1.4. Try another.

    4. Not this one. You stopped at 6 − 2k = 0 and took the 6. Divide by 2 as well: 2k = 6, so k = 3. Go to M5.1.4. Try another.

    See the answer

    3

    Since x + 2 is a factor, f(−2) = 0. f(−2) = −8 + 12 − 2k + 2 = 6 − 2k. So 6 − 2k = 0, and k = 3.

    • If you chose A: You may have written 6 + 2k = 0. The term is kx, and x = −2 gives −2k, so the equation is 6 − 2k = 0. Go to M5.1.4.
    • If you chose C: You put in x = 2. The factor is x + 2, which is x − (−2), so use x = −2. Go to M5.1.4.
    • If you chose D: You stopped at 6 − 2k = 0 and took the 6. Divide by 2 as well: 2k = 6, so k = 3. Go to M5.1.4.

  3. The graph of y = (x − 1)²(x + 2) meets the x axis at two points. What does it do at each?

    Tap your answer.

    1. Not this one. The squared factor is (x − 1)², so the curve touches the axis at x = 1. It crosses where the factor appears once, at x = −2. Go to M5.1.4. Try another.

    2. Not this one. The signs are the wrong way round. The roots come from x − 1 = 0 and x + 2 = 0, giving 1 and −2. Go to M5.1.4. Try another.

    3. Yes, that is right. The factor x + 2 appears once, so the curve crosses at x = −2. The factor (x − 1)² is squared, so the curve touches the axis at x = 1 and turns back.

    4. Not this one. A repeated factor gives one touching point, not two crossings. Here there are only two different points. Go to M5.1.4. Try another.

    See the answer

    It crosses at x = −2 and touches at x = 1.

    The factor x + 2 appears once, so the curve crosses at x = −2. The factor (x − 1)² is squared, so the curve touches the axis at x = 1 and turns back.

    • If you chose A: The squared factor is (x − 1)², so the curve touches the axis at x = 1. It crosses where the factor appears once, at x = −2. Go to M5.1.4.
    • If you chose B: The signs are the wrong way round. The roots come from x − 1 = 0 and x + 2 = 0, giving 1 and −2. Go to M5.1.4.
    • If you chose D: A repeated factor gives one touching point, not two crossings. Here there are only two different points. Go to M5.1.4.

  4. What happens to y = −x³ + 5x as x becomes large and positive?

    Tap your answer.

    1. Yes, that is right. For large x the term −x³ is far bigger than 5x. It is negative when x is positive, so y becomes large and negative. The curve falls to the right.

    2. Not this one. The term with the highest power wins for large x. It is −x³, which is negative when x is positive. Go to M5.1.4. Try another.

    3. Not this one. For large x the cube is much bigger than 5x. At x = 10, x³ is 1000 but 5x is only 50. Go to M5.1.4. Try another.

    4. Not this one. Neither term shrinks as x grows. The powers grow, so y moves far from 0. Go to M5.1.4. Try another.

    See the answer

    y becomes large and negative.

    For large x the term −x³ is far bigger than 5x. It is negative when x is positive, so y becomes large and negative. The curve falls to the right.

    • If you chose B: The term with the highest power wins for large x. It is −x³, which is negative when x is positive. Go to M5.1.4.
    • If you chose C: For large x the cube is much bigger than 5x. At x = 10, x³ is 1000 but 5x is only 50. Go to M5.1.4.
    • If you chose D: Neither term shrinks as x grows. The powers grow, so y moves far from 0. Go to M5.1.4.

  5. Priya finds f(1) = 0, f(−2) = 0 and f(3) = 0 for f(x) = x³ − 2x² − 5x + 6. She writes f(x) = (x + 1)(x − 2)(x + 3). What is wrong?

    Tap your answer.

    1. Not this one. f(1) = 0 gives the factor x − 1, not x + 1. The sign flips. Go to M5.1.4. Try another.

    2. Yes, that is right. A root a gives the factor x − a. The roots 1, −2 and 3 give x − 1, x + 2 and x − 3. Expanding them gives x³ − 2x² − 5x + 6, which is f(x).

    3. Not this one. f(0) is the constant term, the y intercept. It does not give a factor on its own. Go to M5.1.4. Try another.

    See the answer

    Every sign is the wrong way round. The factors are (x − 1)(x + 2)(x − 3).

    A root a gives the factor x − a. The roots 1, −2 and 3 give x − 1, x + 2 and x − 3. Expanding them gives x³ − 2x² − 5x + 6, which is f(x).

    • If you chose A: f(1) = 0 gives the factor x − 1, not x + 1. The sign flips. Go to M5.1.4.
    • If you chose C: f(0) is the constant term, the y intercept. It does not give a factor on its own. Go to M5.1.4.

  6. Solve x³ − 3x² − 6x + 8 = 0. It has exactly one negative root. Find it.

    See the answer

    −2

    Try the divisors of 8 with a minus sign. f(−1) = −1 − 3 + 6 + 8 = 10, which is not 0. f(−2) = −8 − 12 + 12 + 8 = 0. So x + 2 is a factor and the negative root is −2. (The other roots are 1 and 4.)

  7. x + 1 and x + 2 are factors of x³ + px² + qx + 2. Find p and q. Type p first, then q.

    See the answer

    p = 4, q = 5

    f(−1) = 0 gives −1 + p − q + 2 = 0, so p − q = −1. f(−2) = 0 gives −8 + 4p − 2q + 2 = 0, so 2p − q = 3. Subtract the first from the second: p = 4. Then q = p + 1 = 5.

  8. Factorise f(x) = 2x³ − 3x² − 11x + 6 completely. Write the factors with the roots in order, smallest first.

    See the answer

    (x + 2)(2x − 1)(x − 3)

    Try 3: f(3) = 54 − 27 − 33 + 6 = 0, so x − 3 is a factor. (Check the other bracket too: f(1/2) = 1/4 − 3/4 − 11/2 + 6 = 0, so 2x − 1 is a factor.) Dividing gives 2x² + 3x − 2. Find two numbers with product −4 and sum 3: 4 and −1. So 2x² + 4x − x − 2 = 2x(x + 2) − (x + 2) = (2x − 1)(x + 2). So f(x) = (x + 2)(2x − 1)(x − 3).

  9. From an earlier chapter

    What is the remainder when x³ + 2x² − 4x + 5 is divided by x + 3?

    Tap your answer.

    1. Not this one. You worked out f(3). The divisor x + 3 is x − (−3), so the number to put in is −3. Go to M5.1.3. Try another.

    2. Not this one. That is f(0), the constant term. It is the remainder only when you divide by x. Go to M5.1.3. Try another.

    3. Not this one. Check the signs: f(−3) = −27 + 18 + 12 + 5. The term −4x with x = −3 gives +12, not −12. Go to M5.1.3. Try another.

    4. Yes, that is right. The divisor x + 3 is x − (−3), so the remainder is f(−3) = (−3)³ + 2 × (−3)² − 4 × (−3) + 5 = −27 + 18 + 12 + 5 = 8.

    See the answer

    8

    The divisor x + 3 is x − (−3), so the remainder is f(−3) = (−3)³ + 2 × (−3)² − 4 × (−3) + 5 = −27 + 18 + 12 + 5 = 8.

    • If you chose A: You worked out f(3). The divisor x + 3 is x − (−3), so the number to put in is −3. Go to M5.1.3.
    • If you chose B: That is f(0), the constant term. It is the remainder only when you divide by x. Go to M5.1.3.
    • If you chose C: Check the signs: f(−3) = −27 + 18 + 12 + 5. The term −4x with x = −3 gives +12, not −12. Go to M5.1.3.

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.

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

No internet tomorrow? Save the whole book as one file (881 KB). It opens on any phone.