Solving real polynomial equations | Year 12

01 / LEARN

Understand the idea.

Make one side zero before factoring

The zero-product principle applies to a product equal to zero. Move all terms to one side before factoring; then set each factor equal to zero. Dividing by a variable factor can accidentally discard a zero root.

For a biquadratic equation, put $u = x^{2}$ and solve the quadratic in $u$ . Return to the original variable: positive $u$ gives two real square roots, zero gives one, and negative $u$ gives no real $x$ .

Account for every root and every restriction

For a higher-degree equation, remove a known linear factor and solve the remaining polynomial. Check that your list accounts for every factor, including repeated ones; distinct answers need not equal the degree.

In a model, separate algebraic roots from admissible values. A side length must be positive, for example. Reject a root only by citing a stated restriction, and include units with accepted quantities. Do not discard a valid positive root just because another one is simpler.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A quadratic in a square

Solve $x^{4} - 5 x^{2} + 4 = 0$ over the reals.

Worked solution

1
Introduce the squared variable.
$u = x^{2}, u^{2} - 5 u + 4 = 0.$
The original powers are $x^{4} = {(x^{2})}^{2}$ and $x^{2}$ .
2
Solve the quadratic in $u$ .
$(u - 1) (u - 4) = 0 \Rightarrow u = 1 or u = 4.$
The zero-product principle separates the two possible squared values.
3
Return to $x$ and take both signs.
$\begin{aligned} x^{2} = 1 & \Rightarrow x = \pm 1, \\ x^{2} = 4 & \Rightarrow x = \pm 2. \end{aligned}$
Each positive square has two real square roots.
4
List all real solutions.
$x \in {- 2, - 1, 1, 2} .$
The factorisation $(x^{2} - 1) (x^{2} - 4)$ accounts for every root of the quartic.

Answer

$x = - 2, - 1, 1, 2$

EXAMPLE 02

A volume constraint

A cube has side length $s$ cm. Its volume is $6 s^{2} + 7 s - 12$ cubic centimetres. Find all positive side lengths satisfying the model.

Worked solution

1
Equate the two volume expressions.
$s^{3} = 6 s^{2} + 7 s - 12 \Rightarrow s^{3} - 6 s^{2} - 7 s + 12 = 0, s > 0.$
A cube of side $s$ centimetres has volume $s^{3}$ cubic centimetres.
2
Find and remove a linear factor.
$\begin{aligned} P (1) & = 1 - 6 - 7 + 12 = 0, \\ P (s) & = (s - 1) (s^{2} - 5 s - 12) . \end{aligned}$
The factor theorem justifies division by $s - 1$ .
3
Solve the remaining quadratic.
$s = 1 or s = \frac{5 \pm \sqrt{25 + 48}}{2} = \frac{5 \pm \sqrt{73}}{2} .$
The zero-product principle and quadratic formula give all algebraic roots.
4
Apply the positive-length restriction.
$\begin{aligned} \sqrt{73} > 5 & \Rightarrow \frac{5 - \sqrt{73}}{2} < 0, \\ s & = 1 cm or s = \frac{5 + \sqrt{73}}{2} cm . \end{aligned}$
The negative quadratic root is inadmissible, while both positive roots satisfy the stated model.

Answer

$s = 1$ cm or $s = \frac{5 + \sqrt{73}}{2}$ cm.

03 / YOUR TURN

Try it yourself.

Check your understanding

Solve $x^{3} = 4 x$ over the reals.

Show answer & reasoning

$x = - 2, 0, 2$ .

Rearrange to $x^{3} - 4 x = x (x - 2) (x + 2) = 0$ . Setting each factor equal to zero retains all three real roots: $- 2, 0, 2$ .

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Solve $x^{3} - 7 x + 6 = 0$ .
2 marks
Solve $x^{4} - 10 x^{2} + 9 = 0$ .
2 marks
Solve $x^{4} + x^{2} - 6 = 0$ over the reals.
2 marks
Solve $2 x^{3} - 3 x^{2} - 8 x + 12 = 0$ .
2 marks
A model gives $t (t - 2) (t + 5) = 0$ for a time with $0 < t < 4$ . State the allowed solution and explain both exclusions.
2 marks
A student solves $x^{3} = 9 x$ by dividing by $x$ and reports $x = \pm 3$ . Correct the answer.
3 marks

Read the worked solutions

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

Solve $x^{3} - 7 x + 6 = 0$ .
2 marks
Factor as $(x + 3) (x - 1) (x - 2) = 0$ .
$x = - 3, 1, 2$
Solve $x^{4} - 10 x^{2} + 9 = 0$ .
2 marks
Put $u = x^{2}$ ; $(u - 1) (u - 9) = 0$ .
$x = \pm 1, \pm 3$
Solve $x^{4} + x^{2} - 6 = 0$ over the reals.
2 marks
$(x^{2} + 3) (x^{2} - 2) = 0$ ; the first factor has no real zero.
$x = \pm \sqrt{2}$
Solve $2 x^{3} - 3 x^{2} - 8 x + 12 = 0$ .
2 marks
Factor as $(2 x - 3) (x - 2) (x + 2) = 0$ .
$x = - 2, \frac{3}{2}, 2$
A model gives $t (t - 2) (t + 5) = 0$ for a time with $0 < t < 4$ . State the allowed solution and explain both exclusions.
2 marks
$t = 0$ is excluded by the strict lower bound; $t = - 5$ is negative and outside the domain.
$t = 2$ only.
A student solves $x^{3} = 9 x$ by dividing by $x$ and reports $x = \pm 3$ . Correct the answer.
3 marks
The original equation is also true at $x = 0$ . Write $x (x - 3) (x + 3) = 0$ to keep all roots.
$x = - 3, 0, 3$

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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.

Make one side zero before factoring

Account for every root and every restriction

See the method in action.

A quadratic in a square

Introduce the squared variable.

Solve the quadratic in u.

Return to x and take both signs.

List all real solutions.

A volume constraint

Equate the two volume expressions.

Find and remove a linear factor.

Solve the remaining quadratic.

Apply the positive-length restriction.

Try it yourself.

Check your understanding

Keep practising

Solve the quadratic in $u$ .

Return to $x$ and take both signs.