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Open Math›Year 8›Algebraic techniques›Factorising expressions

Factorising expressions

Find the HCF of simple expressions and factorise by taking out a numeric or variable common factor; factorise three-term expressions with a common numeric factor and verify by expanding.

Worked examples

Factorising with a numeric HCF

Straightforward

Problem

Factorise

10x + 15

.

1

Find the HCF of the two coefficients, 10 and 15.

Factors of 10: 1, 2, 5, 10. Factors of 15: 1, 3, 5, 15. The largest common factor is

5

.

2

Write the HCF outside the brackets.

10x + 15 = 5(\ldots)

3

Divide each term by the HCF to find what goes inside the brackets.

\dfrac{10x}{5} = 2x

and

\dfrac{15}{5} = 3

, so

10x + 15 = 5(2x + 3)

4

Verify by expanding.

5(2x + 3) = 10x + 15

✓

Answer

5(2x + 3)

Factorising with a variable HCF

Moderate

Problem

Factorise

6x^2 - 9x

.

1

Find the HCF of the numeric parts: HCF of 6 and 9.

Factors of 6: 1, 2, 3, 6. Factors of 9: 1, 3, 9. Numeric HCF

= 3

.

2

Find the HCF of the variable parts: HCF of

x^2

and

x

.

x^2 = x \times x

and

x = x \times 1

. Variable HCF

= x

.

3

Combine to get the full HCF, then write it outside the brackets.

HCF

= 3x

, so

6x^2 - 9x = 3x(\ldots)

4

Divide each term by

3x

to find what goes inside the brackets, then verify.

\dfrac{6x^2}{3x} = 2x

and

\dfrac{9x}{3x} = 3

, so

6x^2 - 9x = 3x(2x - 3)

. Check:

3x(2x - 3) = 6x^2 - 9x

✓

Answer

3x(2x - 3)

Factorising a three-term expression and verifying

Challenging

Problem

Factorise

6x^2 + 12x + 18

and verify your answer by expanding.

1

Find the HCF of all three coefficients: 6, 12, and 18.

6 = 2 \times 3

,

12 = 2^2 \times 3

,

18 = 2 \times 3^2

. HCF

= 2 \times 3 = 6

.

2

Write the HCF outside the brackets.

6x^2 + 12x + 18 = 6(\ldots)

3

Divide each term by 6 to find what goes inside the brackets.

\dfrac{6x^2}{6} = x^2

,

\quad \dfrac{12x}{6} = 2x

,

\quad \dfrac{18}{6} = 3

4

Write the factorised form, then verify by expanding.

6x^2 + 12x + 18 = 6(x^2 + 2x + 3)

. Check:

6(x^2 + 2x + 3) = 6x^2 + 12x + 18

✓

Answer

6(x^2 + 2x + 3)

Practise

Q1·Straightforward

What is the highest common factor (HCF) of

10x

and

15

?

Explanation

Factors of 10: 1, 2, 5, 10. Factors of 15: 1, 3, 5, 15. The largest factor common to both is 5. So

\text{HCF}(10x, 15) = 5

.

Q2·Straightforward

Factorise

8x + 12

. What is the number that goes outside the brackets?

Explanation

The HCF of 8 and 12 is 4. Dividing each term by 4:

\frac{8x}{4} = 2x

and

\frac{12}{4} = 3

. So

8x + 12 = 4(2x + 3)

. The number outside the brackets is 4.

Q3·Straightforward

Factorise

15y - 20

. What is the coefficient of

y

inside the brackets?

Explanation

The HCF of 15 and 20 is 5. Dividing each term:

\frac{15y}{5} = 3y

and

\frac{20}{5} = 4

. So

15y - 20 = 5(3y - 4)

. The coefficient of

y

inside the brackets is 3.

Q4·Straightforward

Factorise

12a - 18

by taking out the highest common factor.

$6(2a - 3)$$3(4a - 6)$$6(2a + 3)$$2(6a - 9)$

Explanation

The HCF of 12 and 18 is 6. Dividing:

\frac{12a}{6} = 2a

and

\frac{18}{6} = 3

. So

12a - 18 = 6(2a - 3)

. Option B uses 3 (not the HCF) and option D uses 2 (not the HCF) — neither takes out the full HCF. Option C has the wrong sign.

Q5·Moderate

Factorise

4x^2 + 8x

by taking out the HCF. The factorised form is

4x(x + k)

. Find

k

.

Explanation

The HCF of

4x^2

and

8x

is

4x

. Dividing each term:

\frac{4x^2}{4x} = x

and

\frac{8x}{4x} = 2

. So

4x^2 + 8x = 4x(x + 2)

, giving

k = 2

.

Q6·Moderate

Factorise

6x^2 - 9x

. The factorised form is

3x(ax - 3)

. Find

a

.

Explanation

The HCF of

6x^2

and

9x

is

3x

. Dividing:

\frac{6x^2}{3x} = 2x

and

\frac{9x}{3x} = 3

. So

6x^2 - 9x = 3x(2x - 3)

, giving

a = 2

.

Q7·Moderate

Which of the following is the fully factorised form of

10m^2 + 15m

?

$5(2m^2 + 3m)$$5m(2m + 3)$$5m(2m + 4)$$10m(m + 2)$

Explanation

The HCF of

10m^2

and

15m

is

5m

. Dividing:

\frac{10m^2}{5m} = 2m

and

\frac{15m}{5m} = 3

. The fully factorised form is

5m(2m + 3)

. Option A takes out only 5, leaving

m

still common inside the brackets, so it is not fully factorised.

Q8·Moderate

Factorise

12y^2 - 8y

fully. The factorised form is

4y(3y + k)

. Find

k

.

Explanation

The HCF of

12y^2

and

8y

is

4y

. Dividing:

\frac{12y^2}{4y} = 3y

and

\frac{8y}{4y} = 2

. So

12y^2 - 8y = 4y(3y - 2)

. Writing in the form

4y(3y + k)

gives

k = -2

.

Q9·Challenging

Factorise

6x^2 + 12x + 18

by taking out the highest common factor. What is the HCF?

Explanation

The HCF of 6, 12, and 18 is 6. Dividing each term:

\frac{6x^2}{6} = x^2

,

\frac{12x}{6} = 2x

,

\frac{18}{6} = 3

. So

6x^2 + 12x + 18 = 6(x^2 + 2x + 3)

. Verify:

6(x^2 + 2x + 3) = 6x^2 + 12x + 18

✓

Q10·Challenging

Factorise

10a^2 - 15a + 20

. The result is

5(2a^2 + pa + 4)

. Find

p

.

Explanation

The HCF of 10, 15, and 20 is 5. Dividing each term:

\frac{10a^2}{5} = 2a^2

,

\frac{-15a}{5} = -3a

,

\frac{20}{5} = 4

. So

10a^2 - 15a + 20 = 5(2a^2 - 3a + 4)

. Writing in the form

5(2a^2 + pa + 4)

gives

p = -3

.

Q11·Challenging

Which of the following is the fully factorised form of

8x^2 + 4x - 12

?

$4(2x^2 + x - 3)$$2(4x^2 + 2x - 6)$$4(2x^2 + x + 3)$$8(x^2 + x - 4)$

Explanation

The HCF of 8, 4, and 12 is 4. Dividing:

\frac{8x^2}{4} = 2x^2

,

\frac{4x}{4} = x

,

\frac{-12}{4} = -3

. The fully factorised form is

4(2x^2 + x - 3)

. Option B takes out only 2, which is not the HCF. Option C has the wrong sign on the constant. Option D is incorrect.

Q12·Challenging

Factorise

12y^2 - 18y + 6

by taking out the HCF, then verify by expanding. What is the constant term inside the brackets?

Explanation

The HCF of 12, 18, and 6 is 6. Dividing:

\frac{12y^2}{6} = 2y^2

,

\frac{-18y}{6} = -3y

,

\frac{6}{6} = 1

. So

12y^2 - 18y + 6 = 6(2y^2 - 3y + 1)

. Verify:

6(2y^2 - 3y + 1) = 12y^2 - 18y + 6

✓. The constant term inside the brackets is 1.

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