Teach Me Maths

Factor Theorem

How to test linear factors, factorise polynomials, find unknown coefficients and solve cubic equations.

Before you start: this topic uses identities and comparing coefficients to find the remaining factor. Review Identities and Comparing Coefficients if you need a refresher.

Zero output means a factor

For a polynomial f(x), (x-a) is a factor if and only if f(a)=0. Set the proposed factor equal to zero to find the input to test. A non-zero output rules that factor out.

For (x+3), test x=-3. For (2x-5), test x=\frac52. Put negative inputs in brackets when raising them to powers.

Worked example 1

Show that (x-2) is a factor of f(x)=x^3+x^2-5x-2

  1. Solve x-2=0, so test x=2: f(2)=2^3+2^2-5(2)-2

  2. Your turn. Calculate 8+4-10-2

    f(2)=0

  3. Answer: (x-2) is a factor because f(2)=0. State this conclusion explicitly.

Explore roots and factors

Try it: find all three inputs with output zero. A root a gives a factor (x-a); the factor and the root have opposite signs when written this way.

Find the remaining factor

After finding a linear factor of a cubic, divide to obtain a quadratic. You can use polynomial division or compare coefficients in a product. Factorise the quadratic if possible, and expand the final product to check it.

Worked example 2

Given that (x-1) is a factor, factorise x^3-4x^2-7x+10 fully.

  1. Write (x-1)(x^2+Ax+B). Its expansion is x^3+(A-1)x^2+(B-A)x-B

  2. Your turn. Compare x^2 coefficients: A-1=-4. Find A

    A=-3

  3. The constant gives B=-10. Check B-A=-10-(-3)=-7. Thus the quadratic is x^2-3x-10=(x-5)(x+2)
  4. Answer: (x-1)(x-5)(x+2)

Worked example 3

Show that 2x+1 is a factor of 2x^3-3x^2-8x-3, then factorise fully.

  1. Your turn. Which input makes 2x+1=0?

    Test x=-\frac12

  2. f(-\frac12)=-\frac14-\frac34+4-3=0, so 2x+1 is a factor.
  3. Dividing by 2x+1 gives x^2-2x-3=(x-3)(x+1)
  4. Answer: (2x+1)(x-3)(x+1)

Worked example 4

Find k if (x+2) is a factor of x^3+kx^2-3x+6

  1. Test x=-2 and set the result equal to zero: -8+4k+6+6=0

  2. Your turn. Solve 4k+4=0

    k=-1

  3. Answer: k=-1. Checking gives -8-4+6+6=0

Worked example 5

Solve x^3-6x^2+5x+12=0, given that x=3 is a root.

  1. The root gives the factor (x-3). Division gives x^2-3x-4=(x-4)(x+1)

  2. Your turn. What roots come from (x-4)(x+1)=0?

    Give every value, separated by commas

    x=-1 or x=4

  3. Answer: x=-1,3,4. Include the root supplied in the question as well as the two new roots.

Worked example 6

Both (x-1) and (x+2) are factors of x^3+px^2+qx-6. Find p and q.

  1. f(1)=0 gives p+q=5. Also f(-2)=0 gives -8+4p-2q-6=0, so 2p-q=7

  2. Your turn. Add the equations to obtain 3p=12. Find p

    p=4

  3. Then q=5-4=1. The polynomial is (x-1)(x+2)(x+3)
  4. Answer: p=4, q=1

Common mistakes

  • Testing the wrong input. Set the proposed factor to zero; ax+b needs x=-b/a.
  • Leaving the conclusion unstated. Explain that f(r)=0 means (x-r) is a factor.
  • Stopping at a factorisable quadratic. Continue until the expression is fully factorised.
  • Forgetting the supplied root. List every valid root, including any given in the question.

Now try it: Factor Theorem practice questions

More on this topic: Factor Theorem worksheet with full solutions