Solving real polynomial equations | Year 12

01 / LEARN

The idea, step by step

Make one side zero before factoring

The zero-product principle applies when a product is equal to zero: at least one factor must be zero. If the original equation has a non-zero right-hand side, first move everything to one side. Expanding may reveal grouping; keeping a factored form may reveal roots directly.

Look for a substitution such as $u = x^{2}$ in a biquadratic equation. Solve the quadratic in $u$ , then return to $x$ and consider which values of $u$ can equal $x^{2}$ for real $x$ . Do not stop at the substituted variable.

Account for every root and every restriction

After finding a root of a cubic, use division and solve the remaining quadratic. For a graph-based approximation, state the interval and required precision. An approximate root should be checked in the original equation; rounding can change the displayed residual.

Applications often impose positive lengths, limited times or capacity restrictions. Solve the mathematical equation first, then explain which real roots fit that domain. A rejected length is not an algebraic mistake if it is genuinely a root; it is an inadmissible model result.

A common trap

Cancel a factor of x before solving an equation equal to zero.

Instead: Dividing by x loses the possible solution x=0. Factor and apply the zero-product principle instead.

02 / WORKED EXAMPLES

See the thinking

Example 1

A quadratic in a square

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

Let $u = x^{2}$ . Then $u^{2} - 5 u + 4 = (u - 1) (u - 4) = 0$ .
$x^{2} = 1$ gives $x = \pm 1$ ; $x^{2} = 4$ gives $x = \pm 2$ .

x = - 2, - 1, 1, 2

Example 2

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.

$s^{3} = 6 s^{2} + 7 s - 12$ , so $s^{3} - 6 s^{2} - 7 s + 12 = 0$ .
Test $s = 1$ : it is a root. Divide to get $(s - 1) (s^{2} - 5 s - 12) = 0$ .
The quadratic roots are $\frac{5 \pm \sqrt{73}}{2}$ . Both $1$ and $\frac{5 + \sqrt{73}}{2}$ are positive and fit the stated domain; the negative root does not.

s = 1

cm or

s = \frac{5 + \sqrt{73}}{2}

cm.

Check your understanding

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

Show answer

$x = - 2, 0, 2$ .

$x (x^{2} - 4) = 0$ , so retain the zero root as well as $x = \pm 2$ .

03 / WORKSHEET

Your turn

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

Worksheet PDF 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.
$x = - 3, 0, 3$

Checking saved progress…

Ready for the topic test?

Review the chapter, then choose a practice test.

Topic review

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