Finding and factorising real polynomial roots | Year 12

01 / LEARN

Understand the idea.

Reduce the degree one factor at a time

Look first for a common factor, grouping, a difference of squares or a quadratic in $x^{2}$ . When no structure is immediate, test plausible roots using the factor theorem, then divide out a successful linear factor.

For integer coefficients, a rational root in lowest terms $p / q$ must have $p$ dividing the constant coefficient and $q$ dividing the leading coefficient. These are candidates to test, not a guarantee of a rational root.

What complete real factorisation means

Complete factorisation over the reals uses linear factors and quadratic factors with negative discriminant. Such an irreducible quadratic contributes no real zeros, although it still contributes two to the degree.

Account for multiplicities and the leading coefficient. A cubic reduced to a linear factor times a quadratic is finished only after the quadratic has been analysed. Exact substitution verifies roots suggested by a graph; a rounded graph location alone does not establish an exact factor.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A non-monic cubic

Factor $2 x^{3} - 3 x^{2} - 11 x + 6$ over the reals.

Worked solution

1
Test a candidate root.
$P (3) = 2 (27) - 3 (9) - 11 (3) + 6 = 54 - 27 - 33 + 6 = 0.$
The zero remainder proves that $x - 3$ is a factor.
2
Divide by the known factor.
$P (x) = (x - 3) (2 x^{2} + 3 x - 2) .$
The quotient preserves the leading coefficient two and reduces the remaining problem to a quadratic.
3
Factor the quotient by grouping.
$\begin{aligned} 2 x^{2} + 3 x - 2 & = 2 x^{2} + 4 x - x - 2 \\ = 2 x (x + 2) - (x + 2) \\ = (2 x - 1) (x + 2) . \end{aligned}$
Splitting the middle term exposes a common binomial factor.
4
State the full real factorisation.
$P (x) = (x - 3) (2 x - 1) (x + 2) .$
Three linear factors account for the original cubic degree and retain its leading coefficient.

Answer

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

EXAMPLE 02

One real root, not three

Find all real roots of $x^{3} - 2 x^{2} + x - 2 = 0$ .

Worked solution

1
Group terms with a common binomial.
$\begin{aligned} x^{3} - 2 x^{2} + x - 2 & = x^{2} (x - 2) + (x - 2) \\ = (x - 2) (x^{2} + 1) . \end{aligned}$
Both pairs contain the factor $x - 2$ .
2
Apply the zero-product principle.
$x - 2 = 0 or x^{2} + 1 = 0.$
A zero product must contain a zero factor.
3
Determine which factors have real zeros.
$x = 2, x^{2} + 1 \geq 1 > 0 for x \in ℝ .$
The quadratic cannot vanish at any real input, so $2$ is the only real root.

Answer

$x = 2$ only.

03 / YOUR TURN

Try it yourself.

Check your understanding

Factor $x^{3} - 4 x$ and list its real roots.

Show answer & reasoning

$x (x - 2) (x + 2)$ ; roots $- 2, 0, 2$ .

First extract $x$ : $x^{3} - 4 x = x (x^{2} - 4) = x (x - 2) (x + 2)$ . The three real roots are $0, 2, - 2$ , each obtained by setting one factor equal to zero.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Factor $x^{3} - 7 x + 6$ .
2 marks
Factor $x^{3} + x^{2} - 4 x - 4$ .
2 marks
Factor $2 x^{3} - 3 x^{2} - 8 x + 12$ .
2 marks
Factor $x^{4} - 5 x^{2} + 4$ and state all its real roots.
2 marks
Factor $x^{3} - 3 x^{2} - 4 x + 12$ .
2 marks
Factor $x^{4} - 1$ as far as possible over the reals. Why does one quadratic remain?
3 marks

Read the worked solutions

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

Factor $x^{3} - 7 x + 6$ .
2 marks
Test $1$ , divide to obtain $x^{2} + x - 6$ , then factor.
$(x - 1) (x - 2) (x + 3)$
Factor $x^{3} + x^{2} - 4 x - 4$ .
2 marks
Grouping gives $(x + 1) (x^{2} - 4)$ .
$(x + 1) (x - 2) (x + 2)$
Factor $2 x^{3} - 3 x^{2} - 8 x + 12$ .
2 marks
Grouping gives $(2 x - 3) (x^{2} - 4)$ .
$(2 x - 3) (x - 2) (x + 2)$
Factor $x^{4} - 5 x^{2} + 4$ and state all its real roots.
2 marks
Treat the expression as a quadratic in $x^{2}$ : $(x^{2} - 1) (x^{2} - 4)$ .
$(x - 1) (x + 1) (x - 2) (x + 2)$ ; roots $\pm 1, \pm 2$ .
Factor $x^{3} - 3 x^{2} - 4 x + 12$ .
2 marks
Grouping gives $(x - 3) (x^{2} - 4)$ .
$(x - 3) (x - 2) (x + 2)$
Factor $x^{4} - 1$ as far as possible over the reals. Why does one quadratic remain?
3 marks
Use difference of squares twice. The quadratic $x^{2} + 1$ has negative discriminant, so has no real linear factors.
$(x - 1) (x + 1) (x^{2} + 1)$

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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.

Official curriculum source ↗

In this lesson

Understand the idea.

Reduce the degree one factor at a time

What complete real factorisation means

See the method in action.

A non-monic cubic

Test a candidate root.

Divide by the known factor.

Factor the quotient by grouping.

State the full real factorisation.

One real root, not three

Group terms with a common binomial.

Apply the zero-product principle.

Determine which factors have real zeros.

Try it yourself.

Check your understanding

Keep practising