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Year 12 · Real Polynomials · 25 min

Factor theorem

Use the factor theorem to prove a linear factor exists and determine missing coefficients.

01 / LEARN

The idea, step by step

Zero remainder is the key

The factor theorem states that x−a is a factor of P⁢(x) if and only if P⁢(a)=0. It is the zero-remainder case of the remainder theorem. Proving that a proposed factor works is a short calculation followed by a clear conclusion.

When a coefficient is unknown, substitute the zero of the proposed factor and solve the resulting equation. A condition such as “x+2 is a factor” means use x=−2, not x=2. Once the coefficient is found, a quotient gives the remaining factorisation.

A factor does not necessarily determine multiplicity

One successful substitution proves at least one copy of the linear factor. To show that the factor repeats, divide once and test the quotient at the same zero, or factor the quotient completely. Do not infer a double root merely because the graph looks flat in a rough drawing.

It is efficient to test simple candidate roots suggested by a constant term, but trial alone is not a completeness argument. Once division produces a quadratic, solve it; then account for the entire original degree. Always retain the leading coefficient in the final factorisation.

A common trap

Finding one root solves the whole polynomial equation.

Instead: It gives one factor. Divide it out and solve the remaining factor to account for the other roots.

02 / WORKED EXAMPLES

See the thinking

Example 1

Find a coefficient from a factor

Find k so that x−2 is a factor of P⁢(x)=x3+kx2−4⁢x+4. Then divide by that factor.

  1. P⁢(2)=8+4⁢k−8+4=4⁢k+4. Set this equal to zero to obtain k=−1.
  2. The polynomial becomes x3−x2−4⁢x+4=(x−2)⁢(x2+x−2).
  3. Factor the quadratic to get (x−2)⁢(x+2)⁢(x−1).
k=−1; P=(x−2)⁢(x+2)⁢(x−1).
Example 2

A repeated factor

Show that x−2 is a repeated factor of P⁢(x)=x3−3x2+4.

  1. P⁢(2)=0, so x−2 is a factor.
  2. Division gives P⁢(x)=(x−2)⁢(x2−x−2).
  3. The quotient factors as (x−2)⁢(x+1), so P⁢(x)=(x−2)2(x+1).
x−2 occurs twice.

Check your understanding

Is x+3 a factor of x3+3x2+x+3?

Show answer

Yes.

P⁢(−3)=−27+27−3+3=0.

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Read the questions online
  1. Show that x−1 is a factor of x3−4⁢x+3.
    2 marks
  2. Find k if x+2 is a factor of x3+k⁢x+2.
    2 marks
  3. Find k if 2⁢x−1 is a factor of 4x3+k⁢x−1.
    2 marks
  4. Factor x3−4x2−x+4 completely.
    2 marks
  5. Show that x+1 is a repeated factor of x3+x2−x−1.
    2 marks
  6. P⁢(x)=x3+a⁢x+b has factors x−1 and x+2. Find a,b and the third linear factor.
    4 marks
Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

  1. Show that x−1 is a factor of x3−4⁢x+3.
    2 marks

    P⁢(1)=1−4+3=0.

    It is a factor.
  2. Find k if x+2 is a factor of x3+k⁢x+2.
    2 marks

    P⁢(−2)=−8−2⁢k+2=0, so k=−3.

    k=−3
  3. Find k if 2⁢x−1 is a factor of 4x3+k⁢x−1.
    2 marks

    P⁢(12)=12+k2−1=0.

    k=1
  4. Factor x3−4x2−x+4 completely.
    2 marks

    Group: x2(x−4)−(x−4)=(x−4)⁢(x2−1).

    (x−4)⁢(x−1)⁢(x+1)
  5. Show that x+1 is a repeated factor of x3+x2−x−1.
    2 marks

    Grouping gives (x+1)⁢(x2−1), and x2−1=(x+1)⁢(x−1).

    P=(x+1)2(x−1).
  6. P⁢(x)=x3+a⁢x+b has factors x−1 and x+2. Find a,b and the third linear factor.
    4 marks

    1+a+b=0 and −8−2⁢a+b=0. Solving gives a=−3,b=2; P=(x−1)2(x+2).

    a=−3,b=2; third factor x−1.
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Curriculum reference & lesson scope

SACE Board of South Australia

A recognisable real-polynomial chapter supporting Stage 2 Specialist polynomial factorisation and graph analysis; original lessons use current SACE scope and the school chapter structure requested by the user.

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