Zero-product principle and integer-root equations | Year 9

01 / LEARN

The idea, step by step

Why either factor can be zero

If $A B = 0$ , at least one of $A$ and $B$ must be zero. A product of two non-zero real numbers cannot be zero. Therefore $(x - 2) (x + 5) = 0$ gives two cases: $x - 2 = 0$ or $x + 5 = 0$ . The solutions are $2$ and $- 5$ .

Write both cases before solving them. The word “or” matters: a solution only needs to make one factor zero. Substitute each result into the original equation to confirm it works.

Put zero on one side first

The zero-product principle only applies when the product equals zero. For $x^{2} + 3 x = 10$ , first write $x^{2} + 3 x - 10 = 0$ , then factorise. Do not separately set factors equal to $10$ .

An equation such as $x (x - 4) = 0$ has $x = 0$ as a solution. Dividing both sides by $x$ without considering zero would lose that solution. Keep the product form and use the two cases.

A common trap

For $(x - 3) (x + 1) = 8$ , set $x = 3$ or $x = - 1$ .

Instead: The product is not zero. Rearrange to a quadratic equal to zero before factorising.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

Solve $x^{2} - 3 x - 10 = 0$ .

Factorise: $(x - 5) (x + 2) = 0$ .
Set $x - 5 = 0$ or $x + 2 = 0$ .
Check $25 - 15 - 10 = 0$ and $4 + 6 - 10 = 0$ .

x = 5

x = - 2

Example 2

Worked example

Solve $x^{2} = 6 x$ .

Move all terms to one side: $x^{2} - 6 x = 0$ .
Factor out $x$ : $x (x - 6) = 0$ .
Either $x = 0$ or $x - 6 = 0$ .

x = 0

x = 6

Check your understanding

Solve $(x + 2) (x - 7) = 0$ .

Show answer

$x = - 2$ or $x = 7$

Each bracket can independently be zero.

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Solve $(x - 4) (x + 6) = 0$ .
2 marks
Solve $x^{2} + 7 x + 12 = 0$ .
2 marks
Solve $x^{2} - 2 x - 15 = 0$ .
2 marks
Solve $x^{2} = 9 x$ .
2 marks
Solve $x^{2} + x = 20$ .
3 marks
A student divides $x^{2} = 5 x$ by $x$ and gives only $x = 5$ . Find the missing solution and explain.
3 marks

Read the worked solutions

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

Solve $(x - 4) (x + 6) = 0$ .
2 marks
$x - 4 = 0$ or $x + 6 = 0$ .
$x = 4, - 6$
Solve $x^{2} + 7 x + 12 = 0$ .
2 marks
$(x + 3) (x + 4) = 0$ .
$x = - 3, - 4$
Solve $x^{2} - 2 x - 15 = 0$ .
2 marks
$(x - 5) (x + 3) = 0$ .
$x = 5, - 3$
Solve $x^{2} = 9 x$ .
2 marks
$x^{2} - 9 x = x (x - 9) = 0$ .
$x = 0, 9$
Solve $x^{2} + x = 20$ .
3 marks
$x^{2} + x - 20 = (x + 5) (x - 4) = 0$ .
$x = 4, - 5$
A student divides $x^{2} = 5 x$ by $x$ and gives only $x = 5$ . Find the missing solution and explain.
3 marks
$x (x - 5) = 0$ gives $x = 0$ or $5$ .
Division by $x$ wrongly excludes $x = 0$ .
$x = 0$ is also a solution.

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Curriculum reference & lesson scope

Australian Curriculum v9 / SA Curriculum Mathematics

AC9M9A02, AC9M9A04–A06: monic quadratic expressions, integer-root equations, graphs and models.

Official curriculum source ↗

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