Cubic factors and roots | Year 11

01 / LEARN

The idea, step by step

Test a possible root

The factor x-r belongs to f(x) exactly when f(r)=0. Substitute with brackets and include every term. One successful test gives a factor, not automatically all roots.

For a monic polynomial with integer coefficients, any integer root divides the constant. This produces a useful candidate list, although some polynomials have no integer roots.

f (r) = 0 \Leftrightarrow (x - r) is a factor

Reduce the degree

Divide the cubic by the known factor using long division, coefficient comparison or synthetic division. Insert a zero coefficient when a power is missing.

Factor or solve the quotient. Expand the final product as a check. Repeated factors give repeated roots, which matter when sketching the graph.

A common trap

Testing f(r) for a possible factor x+r.

Instead: The factor is zero at x=-r, so test f(-r).

02 / WORKED EXAMPLES

See the thinking

Example 1

Reduce to a quadratic

Factorise $x^{3} - 2 x^{2} - 5 x + 6$ .

Test x=1: 1-2-5+6=0.
Divide by x-1 to get $x^{2} - x - 6$ .
Factor the quotient into $(x - 3) (x + 2)$ .

(x - 1) (x - 3) (x + 2)

Example 2

A coefficient condition

For $f (x) = x^{3} + k x^{2} - 4 x - 4$ , x+1 is a factor. Find k and factorise.

Set f(-1)=-1+k+4-4=k-1=0.
Thus k=1; divide by x+1.
The quotient is $x^{2} - 4$ .

k = 1

;

f (x) = (x + 1) (x - 2) (x + 2)

Check your understanding

Is x-2 a factor of $x^{3} - 3 x^{2} + 4$ ?

Show answer

Yes.

f(2)=8-12+4=0.

03 / WORKSHEET

Your turn

6 questions · 16 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Show x-2 divides $x^{3} - 4 x^{2} + x + 6$ .
2 marks
Divide $x^{3} - 4 x^{2} + x + 6$ by x-2.
3 marks
Solve $x^{3} - 4 x^{2} + x + 6 = 0$ .
3 marks
Find k when x+2 divides $x^{3} + k x + 6$ .
2 marks
Factorise $x^{3} + 2 x^{2} - 9 x - 18$ .
3 marks
Solve $x^{3} - 3 x + 2 = 0$ , stating repeated roots.
3 marks

Read the worked solutions

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

Show x-2 divides $x^{3} - 4 x^{2} + x + 6$ .
2 marks
Substitute: 8-16+2+6=0.
f(2)=0.
Divide $x^{3} - 4 x^{2} + x + 6$ by x-2.
3 marks
Divide term by term.
Expand $(x - 2) (x^{2} - 2 x - 3)$ to check.
$x^{2} - 2 x - 3$
Solve $x^{3} - 4 x^{2} + x + 6 = 0$ .
3 marks
Use $(x - 2) (x^{2} - 2 x - 3)$ .
Factor the quotient as $(x - 3) (x + 1)$ .
$x = - 1, 2, 3$
Find k when x+2 divides $x^{3} + k x + 6$ .
2 marks
Set f(-2)=-8-2k+6=0.
Solve -2k=2.
$k = - 1$
Factorise $x^{3} + 2 x^{2} - 9 x - 18$ .
3 marks
Group $x^{2} (x + 2) - 9 (x + 2)$ .
Factor difference of squares.
$(x + 2) (x - 3) (x + 3)$
Solve $x^{3} - 3 x + 2 = 0$ , stating repeated roots.
3 marks
Test x=1 and divide to obtain $x^{2} + x - 2$ .
The factorisation is $(x + 2) {(x - 1)}^{2}$ .
$x = - 2$ ; x=1 repeated.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics topics 2.1–2.2: quadratic relationships, cubic and quartic polynomials. Original Ascent teaching sequence.

Official curriculum source ↗

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