Factorising into Single Brackets
Question 11 mark
Factorise fully 6x+18.
Hint
Look for the largest number that divides both coefficients.
Worked solution
Factorise fully 6x+18.
- The highest common factor of 6 and 18 is 6.
- 6x\div6=x and 18\div6=3, so the factorisation is 6(x+3).
Question 21 mark
Factorise fully 8y-12.
Hint
The subtraction sign must stay with the second term.
Worked solution
Factorise fully 8y-12.
- The highest common factor is 4.
- Dividing each term by 4 gives 2y and -3. Therefore 8y-12=4(2y-3).
Question 31 mark
Factorise fully ab+ac.
Hint
Identify the letter appearing in every term.
Worked solution
Factorise fully ab+ac.
- Both terms contain the factor a.
- Removing one a from each term leaves b+c, so the answer is a(b+c).
Question 41 mark
Factorise fully t^2+7t.
Hint
A square contains two factors of the letter.
Worked solution
Factorise fully t^2+7t.
- Each term contains t.
- t^2\div t=t and 7t\div t=7, giving t(t+7).
Question 52 marks
Factorise fully 12p^2+18p.
Hint
Find the common numerical factor and common letter factor.
Worked solution
Factorise fully 12p^2+18p.
- The coefficients share 6 and both terms contain p, so take out 6p.
- 12p^2\div6p=2p and 18p\div6p=3.
- The fully factorised expression is 6p(2p+3).
Question 61 mark
Factorise fully 21+14k.
Hint
The constant can be the first term inside the bracket.
Worked solution
Factorise fully 21+14k.
- Both terms are divisible by 7.
- 21\div7=3 and 14k\div7=2k. The answer is 7(3+2k).
Question 72 marks
Factorise fully 15a^2-10a.
Hint
Include the sign when dividing each term by the common factor.
Worked solution
Factorise fully 15a^2-10a.
- The highest common factor is 5a.
- The terms remaining are 3a and -2, so the answer is 5a(3a-2).
Question 81 mark
Factorise fully 18r+30.
Hint
Taking out only 2 or 3 would not finish the factorisation.
Worked solution
Factorise fully 18r+30.
- The highest common factor of 18 and 30 is 6.
- 18r\div6=3r and 30\div6=5.
- The answer is 6(3r+5); no further common factor remains inside.
Question 92 marks
Factorise fully 10xy+15xz.
Hint
Only letters found in every term can be taken out.
Worked solution
Factorise fully 10xy+15xz.
- Both coefficients share 5 and both terms contain x.
- Divide by 5x: the remaining terms are 2y and 3z.
- Hence the answer is 5x(2y+3z).
Question 102 marks
Factorise fully 14m^3-21m^2.
Hint
Use the smallest power appearing in both terms.
Worked solution
Factorise fully 14m^3-21m^2.
- The common numerical factor is 7 and the common letter factor is m^2.
- 14m^3\div7m^2=2m and 21m^2\div7m^2=3.
- The answer is 7m^2(2m-3).
Question 112 marks
Factorise fully 12x+18y-30.
Hint
Check all three terms before choosing the common factor.
Worked solution
Factorise fully 12x+18y-30.
- The three coefficients share the factor 6.
- Dividing gives 2x, 3y and -5.
- The fully factorised expression is 6(2x+3y-5).
Question 122 marks
Factorise fully q^3+4q^2-2q.
Hint
Use a factor shared by every term, including the last.
Worked solution
Factorise fully q^3+4q^2-2q.
- Each term contains at least one q, but the last term has only one.
- Take out q to leave q^2+4q-2.
- The answer is q(q^2+4q-2).
Question 131 mark
Factorise fully -12x-18.
Hint
A negative common factor can make the bracket easier to read.
Worked solution
Factorise fully -12x-18.
- A common factor of -6 makes the terms inside positive.
- (-12x)\div(-6)=2x and (-18)\div(-6)=3.
- So -6(2x+3) is fully factorised. The equivalent form 6(-2x-3) is also correct.
Question 142 marks
Find the missing number.
16x+24=8(2x+\square) for every x.
Hint
Compare the constant terms on the two sides.
Worked solution
Find the missing number.
16x+24=8(2x+\square) for every x.
- The constant term on the right is 8\times\square.
- It must equal 24, so \square=24\div8=3.
- Check: 8(2x+3)=16x+24.
Question 151 mark
Nina factorises 20x^2+30x as 5x(4x+6). Which statement is correct?
Hint
Check both whether it expands correctly and whether it is fully factorised.
Worked solution
Nina factorises 20x^2+30x as 5x(4x+6). Which statement is correct?
- Her expression is not equivalent to the original.
- Her expression is equivalent, but the bracket still has a common factor of 2.
- She should replace x^2 with 2x.
- No letter can be taken out as a factor.
- Expanding Nina’s answer gives the original expression, so it is equivalent.
- However, 4x+6=2(2x+3), so the complete factorisation is 10x(2x+3).
Question 16Challenge2 marks
Factorise fully 12x^3y-18xy^2.
Hint
Take out only the powers shared by both terms.
Worked solution
Factorise fully 12x^3y-18xy^2.
- The numerical common factor is 6. Both terms contain x and y.
- Dividing by 6xy leaves 2x^2 and -3y.
- The answer is 6xy(2x^2-3y). Expanding restores both original terms.
Question 17Challenge3 marks
A rectangle has area (8x^2+20x) cm², where x>0. One side is shown in the diagram.
Find a simplified expression for the length L, in centimetres.
Hint
Find the common factor that matches the given side.
Worked solution
A rectangle has area (8x^2+20x) cm², where x>0. One side is shown in the diagram.
Find a simplified expression for the length L, in centimetres.
- Factorise the area: 8x^2+20x=4x(2x+5).
- Area equals the product of the side lengths. The given side is 4x, so the other side is L=2x+5.
- Check: multiplying 4x by 2x+5 gives the stated area.
Question 18Challenge3 marks
Factorise fully 12a^3b-18a^2b^2+6a^2b.
Hint
Check what remains when the last term is divided by the common factor.
Worked solution
Factorise fully 12a^3b-18a^2b^2+6a^2b.
- The shared factor is 6a^2b.
- The three quotients are 2a, -3b and 1.
- Therefore the answer is 6a^2b(2a-3b+1). The final term inside is 1, not 0.
Question 19Challenge3 marks
The identity Px(3x+Q)=18x^2-30x holds for every x.
Find P.
2 marks
Find Q.
1 mark
Hint
Expand the left side to compare like terms.
Worked solution
The identity Px(3x+Q)=18x^2-30x holds for every x.
Part (a)
Find P.
- The squared term after expansion is 3Px^2.
- Thus 3P=18, so P=6.
Part (b)
Find Q.
- The linear term is PQx. With P=6, its coefficient is 6Q.
- Set 6Q=-30, giving Q=-5.
- Check: 6x(3x-5)=18x^2-30x.
Question 20Challenge3 marks
Factorise fully (12x^2+8x)+(6x^2-2x).
Hint
Simplify the sum before looking for the common factor.
Worked solution
Factorise fully (12x^2+8x)+(6x^2-2x).
- First collect like terms: 12x^2+6x^2=18x^2 and 8x-2x=6x.
- The sum is 18x^2+6x. Its highest common factor is 6x.
- The fully factorised result is 6x(3x+1).