7.3 The Factor Theorem
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Revision notes for Edexcel AS Level Maths 7.3 The Factor Theorem. Open the guide for explanations and worked examples. Written against the Edexcel AS Level Maths (8MA0) specification, so the content matches what's examinable rather than general Maths background.

7.3 The Factor Theorem

What you'll learn

  • How substituting a number into a polynomial can test for a factor.
  • How the factor theorem links factors, roots, and x-intercepts.
  • How to use algebraic division to reduce a cubic to a quadratic.
  • How to factorise and solve cubic equations in exam-style problems.

1. Prerequisites: polynomials and substitution

A lot of this topic is about being organised with substitution. You will often be given a polynomial such as

f(x)=2x3−5x2−4x+3f(x)=2x^3-5x^2-4x+3f(x)=2x3−5x2−4x+3

and asked to test whether a particular bracket is a factor.

Definition

Polynomial

A polynomial is an expression made from powers of a variable with number coefficients, such as x3−4x+7x^3-4x+7x3−4x+7. The highest power tells you its degree. For example, a cubic has degree 3.

When you see f(2)f(2)f(2), it means “put x=2x=2x=2 into the expression for f(x)f(x)f(x)”.

Example

Evaluating a polynomial

For f(x)=3x3−2x2−11x+6f(x)=3x^3-2x^2-11x+6f(x)=3x3−2x2−11x+6, find f(2)f(2)f(2).

  1. Substitute x=2x=2x=2 into every place where xxx appears.

    f(2)=3(2)3−2(2)2−11(2)+6f(2)=3(2)^3-2(2)^2-11(2)+6f(2)=3(2)3−2(2)2−11(2)+6

  2. Work out the powers first.

    f(2)=3(8)−2(4)−22+6f(2)=3(8)-2(4)-22+6f(2)=3(8)−2(4)−22+6

  3. Simplify carefully.

    f(2)=24−8−22+6=0f(2)=24-8-22+6=0f(2)=24−8−22+6=0

  4. So f(2)=0f(2)=0f(2)=0.

Common Mistake

Substituting negatives

If you substitute a negative number, use brackets. For example, when x=−3x=-3x=−3, write (−3)2=9(-3)^2=9(−3)2=9 and (−3)3=−27(-3)^3=-27(−3)3=−27. Missing brackets is one of the quickest ways to lose accuracy.

2. Factors, roots, and the x-axis

A factor is something that divides exactly into an expression. For example, x−2x-2x−2 is a factor of a polynomial if dividing by x−2x-2x−2 leaves no remainder.

A root or solution of f(x)=0f(x)=0f(x)=0 is a value of xxx that makes the polynomial equal zero.

These two ideas are connected:

  • If x=2x=2x=2 is a root, then x−2x-2x−2 is a factor.
  • If x=−3x=-3x=−3 is a root, then x+3x+3x+3 is a factor.

On a graph, roots are the places where the curve crosses or touches the x-axis.

A cubic graph showing that roots are the x-values where the curve meets the x-axis, and each root corresponds to a linear factor.

Cubic graph showing that f(a)=0 means x minus a is a factor

Key Idea

Root to factor

If the root is x=ax=ax=a, the matching factor is x−ax-ax−a. Be especially careful with signs: root x=−3x=-3x=−3 gives factor x+3x+3x+3.

Example

Matching roots and factors

Write down the factor corresponding to each root.

Coefficient comparison expands the known linear factor times an unknown quadratic, then matches the cubic coefficients term by term.

  1. If x=5x=5x=5 is a root, then the factor is x−5x-5x−5.

  2. If x=−4x=-4x=−4 is a root, then the factor is x+4x+4x+4.

  3. If x=12x=\frac{1}{2}x=21​ is a root, then the factor is x−12x-\frac{1}{2}x−21​. You may also see the equivalent integer-coefficient factor 2x−12x-12x−1.

3. The factor theorem

The factor theorem is the main result for this topic.

Definition

The factor theorem

For a polynomial f(x)f(x)f(x), the bracket x−ax-ax−a is a factor of f(x)f(x)f(x) if and only if f(a)=0f(a)=0f(a)=0.

The factor theorem links the three equivalent ideas of a zero value, a root, and a linear factor.

This means you do not need to divide straight away. You can test a possible factor by substitution first.

Example

Showing a bracket is a factor

Show that x+2x+2x+2 is a factor of f(x)=2x3+3x2−8x−12f(x)=2x^3+3x^2-8x-12f(x)=2x3+3x2−8x−12.

  1. Rewrite the bracket in the form x−ax-ax−a. Since x+2=x−(−2)x+2=x-(-2)x+2=x−(−2), we need to test x=−2x=-2x=−2.

  2. Substitute x=−2x=-2x=−2 into f(x)f(x)f(x).

    f(−2)=2(−2)3+3(−2)2−8(−2)−12f(-2)=2(-2)^3+3(-2)^2-8(-2)-12f(−2)=2(−2)3+3(−2)2−8(−2)−12

  3. Simplify.

    f(−2)=2(−8)+3(4)+16−12f(-2)=2(-8)+3(4)+16-12f(−2)=2(−8)+3(4)+16−12

  4. Finish the calculation.

    f(−2)=−16+12+16−12=0f(-2)=-16+12+16-12=0f(−2)=−16+12+16−12=0

  5. Since f(−2)=0f(-2)=0f(−2)=0, by the factor theorem, x+2x+2x+2 is a factor of f(x)f(x)f(x).

Tip

Exam wording

If the question says “use the factor theorem to show”, you must explicitly calculate the relevant value, get zero, and then state the conclusion.

4. Finding a factor by trying possible roots

For cubics with integer coefficients, likely integer roots often come from the factors of the constant term.

For example, if the constant term is 12, possible integer roots include plus or minus 1, 2, 3, 4, 6, and 12.

Possible integer roots of a cubic with integer coefficients come from the positive and negative factors of the constant term.

Common Mistake

Possible roots are only candidates

Factors of the constant term give values worth trying, but they are not guaranteed roots. You still need to test them by substitution.

Example

Finding a factor of a cubic

Let f(x)=x3−4x2−x+4f(x)=x^3-4x^2-x+4f(x)=x3−4x2−x+4. Find a linear factor.

  1. The constant term is 4, so try small factors such as 1, -1, 2, and -2.

  2. Test x=1x=1x=1 first.

    f(1)=13−4(1)2−1+4f(1)=1^3-4(1)^2-1+4f(1)=13−4(1)2−1+4

  3. Simplify.

    f(1)=1−4−1+4=0f(1)=1-4-1+4=0f(1)=1−4−1+4=0

  4. Since f(1)=0f(1)=0f(1)=0, the factor theorem tells us that x−1x-1x−1 is a factor.

5. Algebraic division after finding a factor

Once you know one linear factor of a cubic, you can divide the cubic by that factor. This leaves a quadratic, which you can usually factorise or solve with the quadratic formula.

Key Idea

Cubic to quadratic

The factor theorem finds a linear factor. Algebraic division then reduces a cubic to a quadratic, which is much easier to finish.

After one linear factor is found, a cubic can be reduced to a quadratic and then fully solved or factorised.

You may use long division or coefficient comparison. Coefficient comparison is often quick once the factor is known.

Example

Factorising a cubic completely

Factorise f(x)=2x3−3x2−11x+6f(x)=2x^3-3x^2-11x+6f(x)=2x3−3x2−11x+6 completely.

  1. Try x=2x=2x=2.

    f(2)=2(2)3−3(2)2−11(2)+6f(2)=2(2)^3-3(2)^2-11(2)+6f(2)=2(2)3−3(2)2−11(2)+6

  2. Simplify.

    f(2)=16−12−22+6=−12f(2)=16-12-22+6=-12f(2)=16−12−22+6=−12

  3. This is not zero, so x−2x-2x−2 is not a factor. Now try x=3x=3x=3.

    f(3)=2(3)3−3(3)2−11(3)+6f(3)=2(3)^3-3(3)^2-11(3)+6f(3)=2(3)3−3(3)2−11(3)+6

  4. Simplify.

    f(3)=54−27−33+6=0f(3)=54-27-33+6=0f(3)=54−27−33+6=0

  5. Therefore x−3x-3x−3 is a factor. Write

    2x3−3x2−11x+6=(x−3)(Ax2+Bx+C)2x^3-3x^2-11x+6=(x-3)(Ax^2+Bx+C)2x3−3x2−11x+6=(x−3)(Ax2+Bx+C)

  6. Expand the right-hand side.

    (x−3)(Ax2+Bx+C)=Ax3+(B−3A)x2+(C−3B)x−3C(x-3)(Ax^2+Bx+C)=Ax^3+(B-3A)x^2+(C-3B)x-3C(x−3)(Ax2+Bx+C)=Ax3+(B−3A)x2+(C−3B)x−3C

  7. Compare coefficients with 2x3−3x2−11x+62x^3-3x^2-11x+62x3−3x2−11x+6.

    A=2,B−3A=−3,C−3B=−11,−3C=6A=2,\qquad B-3A=-3,\qquad C-3B=-11,\qquad -3C=6A=2,B−3A=−3,C−3B=−11,−3C=6

  8. Solve these values. From A=2A=2A=2, then B−6=−3B-6=-3B−6=−3, so B=3B=3B=3. From −3C=6-3C=6−3C=6, C=−2C=-2C=−2. The quadratic factor is 2x2+3x−22x^2+3x-22x2+3x−2.

  9. Factorise the quadratic.

    2x2+3x−2=(2x−1)(x+2)2x^2+3x-2=(2x-1)(x+2)2x2+3x−2=(2x−1)(x+2)

  10. So the complete factorisation is

    f(x)=(x−3)(2x−1)(x+2)f(x)=(x-3)(2x-1)(x+2)f(x)=(x−3)(2x−1)(x+2)
Common Mistake

Stopping too early

If the question says “factorise completely”, do not stop at a linear factor times a quadratic unless the quadratic cannot be factorised further over the required number system.

6. Finding unknown coefficients using given values

Sometimes the polynomial contains unknown constants, such as ppp, qqq, or kkk. Given information like f(4)=0f(4)=0f(4)=0 can be turned into equations.

Example

Finding unknowns, then factorising

Let f(x)=x3+2x2+px+qf(x)=x^3+2x^2+px+qf(x)=x3+2x2+px+q. Given that f(2)=0f(2)=0f(2)=0 and f(−3)=10f(-3)=10f(−3)=10, find ppp and qqq, then factorise f(x)f(x)f(x).

  1. Use f(2)=0f(2)=0f(2)=0.

    23+2(2)2+2p+q=02^3+2(2)^2+2p+q=023+2(2)2+2p+q=0

  2. Simplify to get the first equation.

    8+8+2p+q=08+8+2p+q=08+8+2p+q=0

  3. So

    2p+q=−162p+q=-162p+q=−16

  4. Use f(−3)=10f(-3)=10f(−3)=10.

    (−3)3+2(−3)2−3p+q=10(-3)^3+2(-3)^2-3p+q=10(−3)3+2(−3)2−3p+q=10

  5. Simplify to get the second equation.

    −27+18−3p+q=10-27+18-3p+q=10−27+18−3p+q=10

  6. So

    −3p+q=19-3p+q=19−3p+q=19

  7. Solve the simultaneous equations.

    2p+q=−16−3p+q=19\begin{aligned} 2p+q&=-16\\ -3p+q&=19 \end{aligned}2p+q−3p+q​=−16=19​

  8. Subtract the second equation from the first.

    5p=−355p=-355p=−35

  9. Hence p=−7p=-7p=−7. Substitute into 2p+q=−162p+q=-162p+q=−16.

    2(−7)+q=−162(-7)+q=-162(−7)+q=−16

  10. So q=−2q=-2q=−2, and therefore

    f(x)=x3+2x2−7x−2f(x)=x^3+2x^2-7x-2f(x)=x3+2x2−7x−2

  11. Since f(2)=0f(2)=0f(2)=0, x−2x-2x−2 is a factor. Divide or compare coefficients:

    x3+2x2−7x−2=(x−2)(x2+4x+1)x^3+2x^2-7x-2=(x-2)(x^2+4x+1)x3+2x2−7x−2=(x−2)(x2+4x+1)

  12. The quadratic does not factorise neatly using integers, so the factorised form is

    (x−2)(x2+4x+1)(x-2)(x^2+4x+1)(x−2)(x2+4x+1)

7. Solving cubic equations

To solve a cubic equation:

Solving a cubic follows a standard sequence: find one root, divide, solve the resulting quadratic, then list all roots.

  1. Find one root using the factor theorem.
  2. Divide by the matching linear factor.
  3. Solve the remaining quadratic.

If the quadratic does not factorise, use the quadratic formula:

x=−b±b2−4ac2ax=\frac{-b\pm\sqrt{b^2-4ac}}{2a}x=2a−b±b2−4ac​​
Example

Solving a cubic equation

Solve x3+4x2−x−4=0x^3+4x^2-x-4=0x3+4x2−x−4=0.

  1. Let f(x)=x3+4x2−x−4f(x)=x^3+4x^2-x-4f(x)=x3+4x2−x−4. Test x=1x=1x=1.

    f(1)=1+4−1−4=0f(1)=1+4-1-4=0f(1)=1+4−1−4=0

  2. So x=1x=1x=1 is a solution and x−1x-1x−1 is a factor.

  3. Divide by x−1x-1x−1 or compare coefficients.

    x3+4x2−x−4=(x−1)(x2+5x+4)x^3+4x^2-x-4=(x-1)(x^2+5x+4)x3+4x2−x−4=(x−1)(x2+5x+4)

  4. Factorise the quadratic.

    x2+5x+4=(x+1)(x+4)x^2+5x+4=(x+1)(x+4)x2+5x+4=(x+1)(x+4)

  5. Therefore

    x3+4x2−x−4=(x−1)(x+1)(x+4)x^3+4x^2-x-4=(x-1)(x+1)(x+4)x3+4x2−x−4=(x−1)(x+1)(x+4)

  6. Set each factor equal to zero.

    x−1=0,x+1=0,x+4=0x-1=0,\qquad x+1=0,\qquad x+4=0x−1=0,x+1=0,x+4=0

  7. The solutions are

    x=1,x=−1,x=−4x=1,\qquad x=-1,\qquad x=-4x=1,x=−1,x=−4
Example

When the quadratic needs decimals

Solve x3+5x2+x−7=0x^3+5x^2+x-7=0x3+5x2+x−7=0, giving non-integer answers to 2 decimal places.

  1. Test x=1x=1x=1.

    f(1)=1+5+1−7=0f(1)=1+5+1-7=0f(1)=1+5+1−7=0

  2. So x−1x-1x−1 is a factor.

  3. Divide the cubic by x−1x-1x−1.

    x3+5x2+x−7=(x−1)(x2+6x+7)x^3+5x^2+x-7=(x-1)(x^2+6x+7)x3+5x2+x−7=(x−1)(x2+6x+7)

  4. Solve the quadratic x2+6x+7=0x^2+6x+7=0x2+6x+7=0 using the quadratic formula, where a=1a=1a=1, b=6b=6b=6, and c=7c=7c=7.

    x=−6±62−4(1)(7)2(1)x=\frac{-6\pm\sqrt{6^2-4(1)(7)}}{2(1)}x=2(1)−6±62−4(1)(7)​​

  5. Simplify.

    x=−6±82x=\frac{-6\pm\sqrt{8}}{2}x=2−6±8​​

  6. This gives the two decimal roots.

    x≈−1.59,x≈−4.41x\approx -1.59,\qquad x\approx -4.41x≈−1.59,x≈−4.41

  7. Include the root from the linear factor as well.

    x=1,x≈−1.59,x≈−4.41x=1,\qquad x\approx -1.59,\qquad x\approx -4.41x=1,x≈−1.59,x≈−4.41
Exam technique

In the exam

  1. If asked to “show” a factor, substitute the correct value and clearly write that the result is zero.
  2. After finding a factor of a cubic, divide to get a quadratic, then factorise or use the quadratic formula.
  3. Always check signs: x+3x+3x+3 means test x=−3x=-3x=−3, while x−3x-3x−3 means test x=3x=3x=3.
Self review

Check yourself

  • If f(−4)=0f(-4)=0f(−4)=0, what linear factor must f(x)f(x)f(x) have?
  • How would you show that x−2x-2x−2 is a factor without doing division?
  • After dividing a cubic by a known factor, what type of expression should be left?

Recap questions

Test yourself with 5 quick questions on this guide. Answer them all correctly to complete it.

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