Factor theorem | Year 12 | Ascent Library

01 / LEARN

Understand the idea.

Zero remainder is the key

The factor theorem says that $x - a$ divides $P (x)$ exactly when $P (a) = 0$ . It follows from the remainder theorem: a linear factor means the remainder at its zero vanishes.

To determine an unknown coefficient, substitute the zero of the proposed factor and set the result equal to zero. For a factor $x + 2$ , that input is $- 2$ . Solve the coefficient equation before dividing the resulting polynomial.

A factor does not necessarily determine multiplicity

A single zero substitution proves at least one copy of a factor. To establish a repeated factor, divide once and show the quotient has the same zero, or factor that quotient completely.

Keep reducing the degree until every remaining factor has been accounted for. Preserve the leading coefficient, and verify by multiplication if necessary. A rough graph can suggest a repeated root, but the algebra must establish its multiplicity.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Find a coefficient from a factor

Find $k$ so that $x - 2$ is a factor of $P (x) = x^{3} + k x^{2} - 4 x + 4$ . Then divide by that factor.

Worked solution

1
Impose the factor condition.
$P (2) = 2^{3} + k (2^{2}) - 4 (2) + 4 = 4 k + 4 = 0.$
The factor $x - 2$ requires its zero to be a root of $P$ .
2
Solve for the coefficient.
$4 k = - 4 \Rightarrow k = - 1.$
This is the unique value satisfying the factor condition.
3
Divide the resulting cubic.
$x^{3} - x^{2} - 4 x + 4 = (x - 2) (x^{2} + x - 2) .$
Polynomial division removes the known linear factor with zero remainder.
4
Factor the quadratic quotient.
$\begin{aligned} x^{2} + x - 2 & = (x + 2) (x - 1), \\ P (x) & = (x - 2) (x + 2) (x - 1) . \end{aligned}$
The two constants have product $- 2$ and sum $1$ , completing the factorisation.

Answer

$k = - 1$ ; $P = (x - 2) (x + 2) (x - 1)$ .

EXAMPLE 02

A repeated factor

Show that $x - 2$ is a repeated factor of $P (x) = x^{3} - 3 x^{2} + 4$ .

Worked solution

1
Prove that the first factor occurs.
$P (2) = 8 - 12 + 4 = 0.$
The factor theorem gives one copy of $x - 2$ .
2
Divide out that copy.
$P (x) = (x - 2) (x^{2} - x - 2) .$
The quotient has degree two and the remainder is zero.
3
Find the repeated factor in the quotient.
$x^{2} - x - 2 = (x - 2) (x + 1) .$
The quotient has the same zero $2$ , producing a second copy.
4
State the exact multiplicity.
$P (x) = {(x - 2)}^{2} (x + 1), 2 + 1 = 3 \neq 0.$
The remaining factor is nonzero at $2$ , so $x - 2$ occurs exactly twice.

Answer

$x - 2$ occurs twice.

03 / YOUR TURN

Try it yourself.

Check your understanding

Is $x + 3$ a factor of $x^{3} + 3 x^{2} + x + 3$ ?

Show answer & reasoning

Yes.

The candidate factor $x + 3$ has zero $- 3$ . Substituting gives $P (- 3) = - 27 + 27 - 3 + 3 = 0$ , so the factor theorem confirms that it is a factor.

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

Show that $x - 1$ is a factor of $x^{3} - 4 x + 3$ .
2 marks
Find $k$ if $x + 2$ is a factor of $x^{3} + k x + 2$ .
2 marks
Find $k$ if $2 x - 1$ is a factor of $4 x^{3} + k x - 1$ .
2 marks
Factor $x^{3} - 4 x^{2} - x + 4$ completely.
2 marks
Show that $x + 1$ is a repeated factor of $x^{3} + x^{2} - x - 1$ .
2 marks
$P (x) = x^{3} + a x + b$ 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.

Show that $x - 1$ is a factor of $x^{3} - 4 x + 3$ .
2 marks
$P (1) = 1 - 4 + 3 = 0$ .
It is a factor.
Find $k$ if $x + 2$ is a factor of $x^{3} + k x + 2$ .
2 marks
$P (- 2) = - 8 - 2 k + 2 = 0$ , so $k = - 3$ .
$k = - 3$
Find $k$ if $2 x - 1$ is a factor of $4 x^{3} + k x - 1$ .
2 marks
$P (\frac{1}{2}) = \frac{1}{2} + \frac{k}{2} - 1 = 0$ .
$k = 1$
Factor $x^{3} - 4 x^{2} - x + 4$ completely.
2 marks
Group: $x^{2} (x - 4) - (x - 4) = (x - 4) (x^{2} - 1)$ .
$(x - 4) (x - 1) (x + 1)$
Show that $x + 1$ is a repeated factor of $x^{3} + x^{2} - x - 1$ .
2 marks
Grouping gives $(x + 1) (x^{2} - 1)$ , and $x^{2} - 1 = (x + 1) (x - 1)$ .
$P = {(x + 1)}^{2} (x - 1)$ .
$P (x) = x^{3} + a x + b$ 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 $x - 1$ .

Checking saved progress…

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.

Official curriculum source ↗

In this lesson

Understand the idea.

Zero remainder is the key

A factor does not necessarily determine multiplicity

See the method in action.

Find a coefficient from a factor

Impose the factor condition.

Solve for the coefficient.

Divide the resulting cubic.

Factor the quadratic quotient.

A repeated factor

Prove that the first factor occurs.

Divide out that copy.

Find the repeated factor in the quotient.

State the exact multiplicity.

Try it yourself.

Check your understanding

Keep practising