Solving by factorisation | Year 11

01 / LEARN

Understand the idea.

Zero is essential

The zero-product principle says that a product equals zero exactly when at least one factor equals zero. Rearrange a quadratic equation so one side is zero, factor the other side, and solve every factor equation.

A product equal to a nonzero number does not allow this step: $(x - 2) (x + 5) = 6$ does not mean either factor is six or zero. Always establish the zero on one side before using factorisation to find solutions.

A B = 0 \Rightarrow A = 0 or B = 0

Do not lose a factor

Avoid dividing by an expression that could be zero. For $x^{2} = 4 x$ , division by $x$ discards the possibility $x = 0$ . Moving all terms to one side and factoring $x (x - 4) = 0$ preserves both possibilities.

A repeated factor gives a repeated root, not a second distinct value: ${(x - 3)}^{2} = 0$ has only $x = 3$ . Check solutions in the original equation and apply any restrictions from a context, such as positive lengths.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Rearrange first

Solve $x^{2} + 2 x = 15$ .

Worked solution

1
Rearrange to zero.
$x^{2} + 2 x - 15 = 0$
Subtracting fifteen from both sides makes the zero-product principle available after factorisation.
2
Factor the quadratic.
$(x + 5) (x - 3) = 0$
Five and negative three have sum two and product negative fifteen.
3
Solve each factor equation.
$\begin{aligned} x + 5 = 0 & ⟹ x = - 5 \\ x - 3 = 0 & ⟹ x = 3 \end{aligned}$
Either factor may be zero, so both branches must be retained.
4
Check both original equations.
$\begin{aligned} {(- 5)}^{2} + 2 (- 5) & = 25 - 10 = 15 \\ 3^{2} + 2 (3) & = 9 + 6 = 15 \end{aligned}$
Both candidate inputs produce the required right-hand side, and the factorisation accounts for all roots.

Answer

$x = - 5$ or $x = 3$

EXAMPLE 02

Keep zero

Solve $3 x^{2} = 6 x$ .

Worked solution

1
Move all terms to one side.
$3 x^{2} - 6 x = 0$
Subtraction preserves the zero solution that division by x could lose.
2
Take out the common factor.
$3 x (x - 2) = 0$
Every term contains the factor three x.
3
Solve the variable factors.
$\begin{aligned} x = 0 & or x - 2 = 0 \\ x = 0 & or x = 2 \end{aligned}$
The constant factor three is nonzero; the two variable factors supply the solutions.
4
Check the retained branches.
$3 {(0)}^{2} = 6 (0) = 0, 3 {(2)}^{2} = 6 (2) = 12$
Zero and two both satisfy the original equation.

Answer

$x = 0$ or $x = 2$

03 / YOUR TURN

Try it yourself.

Check your understanding

Solve $(2 x + 1) (x - 6) = 0$ .

Show answer & reasoning

$x = - \frac{1}{2}$ or $x = 6$

The product is zero, so $2 x + 1 = 0$ or $x - 6 = 0$ . These give $x = - \frac{1}{2}$ and $x = 6$ , respectively.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Solve $x^{2} - 9 x + 20 = 0$ .
2 marks
Solve $2 x^{2} + 7 x + 3 = 0$ .
2 marks
Solve $x^{2} = 7 x$ .
2 marks
Solve $4 x^{2} - 25 = 0$ .
2 marks
Solve $x^{2} + 6 x + 9 = 0$ .
2 marks
A rectangle has sides $x$ cm and $(x + 2)$ cm and area 48 square centimetres. Find $x$ .
3 marks

Read the worked solutions

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

Solve $x^{2} - 9 x + 20 = 0$ .
2 marks
Factor $(x - 4) (x - 5) = 0$ .
Use each factor.
$x = 4, 5$
Solve $2 x^{2} + 7 x + 3 = 0$ .
2 marks
Factor $(2 x + 1) (x + 3) = 0$ .
Solve both equations.
$x = - 3, - \frac{1}{2}$
Solve $x^{2} = 7 x$ .
2 marks
Rearrange and factor $x (x - 7) = 0$ .
Keep both factors.
$x = 0, 7$
Solve $4 x^{2} - 25 = 0$ .
2 marks
Factor $(2 x - 5) (2 x + 5) = 0$ .
Solve both factors.
$x = \pm \frac{5}{2}$
Solve $x^{2} + 6 x + 9 = 0$ .
2 marks
Factor ${(x + 3)}^{2} = 0$ .
There is one distinct root.
$x = - 3$ , repeated.
A rectangle has sides $x$ cm and $(x + 2)$ cm and area 48 square centimetres. Find $x$ .
3 marks
Write $x (x + 2) = 48$ , so $(x + 8) (x - 6) = 0$ .
Reject the negative length -8.
$x = 6$ cm

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics topics 2.1–2.2: quadratic relationships, cubic and quartic polynomials. Original Ascent teaching sequence.

Official curriculum source ↗

In this lesson

Understand the idea.

Zero is essential

Do not lose a factor

See the method in action.

Rearrange first

Rearrange to zero.

Factor the quadratic.

Solve each factor equation.

Check both original equations.

Keep zero

Move all terms to one side.

Take out the common factor.

Solve the variable factors.

Check the retained branches.

Try it yourself.

Check your understanding

Keep practising