Finding and factorising real polynomial roots | Year 12

01 / LEARN

The idea, step by step

Reduce the degree one factor at a time

Begin by looking for a common factor, grouping, a difference of squares or a repeated quadratic structure. If these do not immediately work, test plausible rational roots with the factor theorem. For integer coefficients, an integer root must divide the constant term; rational candidates also depend on the leading coefficient.

A successful test gives a factor, not the final answer. Divide out that factor and solve the lower-degree polynomial. A cubic becomes a quadratic after one linear factor is removed. Check every claimed root in the original polynomial and retain the original leading coefficient.

What complete real factorisation means

Over the real numbers, a polynomial may factor into linear factors and irreducible quadratic factors. A quadratic with negative discriminant has no real root, so it remains a quadratic in a real factorisation. This does not mean the polynomial has no complex roots; complex factorisation is a separate chapter.

Account for degree and multiplicity. If a cubic has been written as a non-zero constant times three linear factors, all its roots have been found. An approximate root from a graph may suggest where to search, but an exact factorisation needs algebraic justification.

A common trap

A real cubic must have three distinct real roots.

Instead: It has three complex roots counting multiplicity, but may have only one distinct real root.

02 / WORKED EXAMPLES

See the thinking

Example 1

A non-monic cubic

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

Test $x = 3$ : $54 - 27 - 33 + 6 = 0$ , so $x - 3$ is a factor.
Divide to obtain $(x - 3) (2 x^{2} + 3 x - 2)$ .
The quadratic factors as $(2 x - 1) (x + 2)$ .

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

Example 2

One real root, not three

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

Group: $x^{2} (x - 2) + (x - 2) = (x - 2) (x^{2} + 1)$ .
$x^{2} + 1$ is positive for every real $x$ , so it contributes no real roots.

x = 2

only.

Check your understanding

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

Show answer

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

Extract x, then use the difference of squares.

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF 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.
$(x - 1) (x + 1) (x^{2} + 1)$

Checking saved progress…

Continue learning

Graphing real polynomials

Next lesson

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 ↗

Back to Real Polynomials