Cubic factors and roots | Year 11

01 / LEARN

Understand the idea.

Test a possible root

The factor theorem says $x - r$ is a factor of a polynomial $f (x)$ exactly when $f (r) = 0$ . Substitute the candidate input with brackets and include every term. A successful test identifies one factor; it need not identify all roots.

For a monic polynomial with integer coefficients, any integer root divides the constant term. This gives a finite candidate list, including negative divisors. It does not guarantee an integer root exists, and roots may also be irrational or non-real.

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

Reduce the degree

After finding a cubic’s linear factor, divide it out to obtain a quadratic. Use long division, synthetic division or coefficient comparison, and include zero coefficients for missing powers so that terms stay aligned.

Factor or solve the quadratic quotient, then retain every resulting root and its multiplicity. Check the complete factorisation by expansion. A repeated factor still counts when describing how the graph behaves at its intercept.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Reduce to a quadratic

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

Worked solution

1
Test a divisor of the constant term.
$f (1) = 1 - 2 - 5 + 6 = 0$
The factor theorem identifies x minus one as a factor.
2
Compare coefficients for the quadratic quotient.
$\begin{aligned} (x - 1) (x^{2} + A x + B) & = x^{3} + (A - 1) x^{2} + (B - A) x - B \\ A - 1 & = - 2 \\ B - A & = - 5 \end{aligned}$
Multiplying an unknown monic quadratic by the known factor must reproduce the original coefficients.
3
Determine and factor the quotient.
$A = - 1, B = - 6, x^{2} - x - 6 = (x - 3) (x + 2)$
The constant check gives negative B equal to six; negative three and two have the required sum and product.
4
Write and verify the full factorisation.
$\begin{aligned} f (x) & = (x - 1) (x - 3) (x + 2) \\ (x - 1) (x^{2} - x - 6) & = x^{3} - 2 x^{2} - 5 x + 6 \end{aligned}$
Expansion recovers all four original coefficients, confirming the product.

Answer

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

EXAMPLE 02

A coefficient condition

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

Worked solution

1
Translate the factor condition into a zero.
$f (- 1) = {(- 1)}^{3} + k {(- 1)}^{2} - 4 (- 1) - 4 = k - 1 = 0$
The factor x plus one vanishes at negative one, so the polynomial must vanish there too.
2
Determine the coefficient and rewrite the polynomial.
$k = 1, f (x) = x^{3} + x^{2} - 4 x - 4$
Solving the factor condition fixes the parameter before the remaining factorisation.
3
Group terms to expose the known factor.
$\begin{aligned} f (x) & = x^{2} (x + 1) - 4 (x + 1) \\ = (x + 1) (x^{2} - 4) \end{aligned}$
Both groups contain the complete factor x plus one, leaving the quadratic quotient x squared minus four.
4
Factor the difference of squares.
$f (x) = (x + 1) (x - 2) (x + 2)$
The quotient is the difference of the squares of x and two, so it splits into the two remaining linear factors.

Answer

$k = 1$ ; $f (x) = (x + 1) (x - 2) (x + 2)$

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

Yes.

For $f (x) = x^{3} - 3 x^{2} + 4$ , substitution gives $f (2) = 8 - 12 + 4 = 0$ . The factor theorem therefore confirms that $x - 2$ is a factor.

Keep practising

6 worksheet questions · 16 marks

Open worksheet 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 $f (- 2) = - 8 - 2 k + 6 = 0$ .
Solve $- 2 k = 2$ .
$k = - 1$
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 ↗

In this lesson

Understand the idea.

Test a possible root

Reduce the degree

See the method in action.

Reduce to a quadratic

Test a divisor of the constant term.

Compare coefficients for the quadratic quotient.

Determine and factor the quotient.

Write and verify the full factorisation.

A coefficient condition

Translate the factor condition into a zero.

Determine the coefficient and rewrite the polynomial.

Group terms to expose the known factor.

Factor the difference of squares.

Try it yourself.

Check your understanding

Keep practising