Exam questions · Maths · Algebra
Factorising Expressions
- 7 exam questions
- 16 marks
- 10 quick checks
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1 Factorise [2 marks]
Factorise \(6x + 15\).
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Model answer
The highest common factor of 6 and 15 is 3, so \(6x + 15 = 3(2x + 5)\).
Mark scheme
- \(3(\ldots)\) or a common factor taken out correctly — M1
- \(3(2x + 5)\) — A1
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2 Factorise [3 marks]
Factorise fully (a) \(12a + 18\) (1 mark) (b) \(6x^2y - 9xy^2\) (2 marks)
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Model answer
(a) The HCF of 12 and 18 is 6, so \(6(2a + 3)\). (b) The HCF of the numbers is 3 and both terms contain \(x\) and \(y\), so \(3xy(2x - 3y)\).
Mark scheme
- (a) \(6(2a + 3)\) — B1
- (b) A correct partial factorisation, such as \(3(2x^2y - 3xy^2)\) or \(xy(6x - 9y)\) — M1
- (b) \(3xy(2x - 3y)\) — A1
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3 Factorise [2 marks]
Factorise \(x^2 + 8x + 15\).
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Model answer
The numbers that multiply to give 15 and add to give 8 are 3 and 5, so \(x^2 + 8x + 15 = (x + 3)(x + 5)\).
Mark scheme
- \((x + a)(x + b)\) with \(ab = 15\) or \(a + b = 8\) — M1
- \((x + 3)(x + 5)\) — A1
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4 Factorise [2 marks]
Factorise \(x^2 - 2x - 15\).
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Model answer
The numbers that multiply to give \(-15\) and add to give \(-2\) are \(-5\) and \(3\), so \(x^2 - 2x - 15 = (x - 5)(x + 3)\).
Mark scheme
- \((x + a)(x + b)\) with \(ab = -15\) or \(a + b = -2\) — M1
- \((x - 5)(x + 3)\) — A1
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5 Factorise [3 marks]
(a) Factorise \(x^2 - 81\). (1 mark) (b) Factorise fully \(2y^2 - 50\). (2 marks)
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Model answer
(a) This is a difference of two squares: \(x^2 - 9^2 = (x + 9)(x - 9)\). (b) Take out the common factor 2 first: \(2y^2 - 50 = 2(y^2 - 25) = 2(y + 5)(y - 5)\).
Mark scheme
- (a) \((x + 9)(x - 9)\) — B1
- (b) \(2(y^2 - 25)\) — M1
- (b) \(2(y + 5)(y - 5)\) — A1
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6 Factorise [2 marks]
Factorise \(3x^2 - 10x - 8\).
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Model answer
\(3 \times (-8) = -24\), and \(-12\) and \(2\) multiply to \(-24\) and add to \(-10\). So \(3x^2 - 12x + 2x - 8 = 3x(x - 4) + 2(x - 4) = (3x + 2)(x - 4)\).
Mark scheme
- \((3x + a)(x + b)\) with \(ab = -8\) or the middle term correctly split as \(-12x + 2x\) — M1
- \((3x + 2)(x - 4)\) — A1
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7 Work out [2 marks]
Work out the value of \(101^2 - 99^2\). You must show your working and you must not use a calculator.
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Model answer
Using the difference of two squares, \(101^2 - 99^2 = (101 + 99)(101 - 99) = 200 \times 2 = 400\).
Mark scheme
- \((101 + 99)(101 - 99)\) — M1
- 400 — A1
Quick check
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1
Factorise \(x^2 - 10x + 25\).
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A: \((x - 5)^2\)
The numbers are \(-5\) and \(-5\), so this is the perfect square \((x - 5)^2\).
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2
Simplify \(\dfrac{x^2 - 9}{x + 3}\).
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A: \(x - 3\)
\(x^2 - 9 = (x + 3)(x - 3)\), so the \((x + 3)\) cancels, leaving \(x - 3\).
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3
Factorise \(6x + 15\).
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D: \(3(2x + 5)\)
The highest common factor of 6 and 15 is 3, and \(3(2x + 5) = 6x + 15\).
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4
Factorise fully \(8x^2 - 12x\).
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C: \(4x(2x - 3)\)
The HCF of 8 and 12 is 4, and \(x\) is in both terms, so \(4x\) comes out: \(4x(2x - 3)\).
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5
Which expression expands to \(x^2 + x - 12\)?
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A: \((x + 4)(x - 3)\)
\((x + 4)(x - 3) = x^2 - 3x + 4x - 12 = x^2 + x - 12\).
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6
Factorise \(x^2 + 9x + 18\).
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B: \((x + 3)(x + 6)\)
The numbers that multiply to 18 and add to 9 are 3 and 6.
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7
Factorise \(x^2 - 36\).
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B: \((x + 6)(x - 6)\)
This is the difference of two squares: \(x^2 - 6^2 = (x + 6)(x - 6)\).
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8
Which of these cannot be factorised as a difference of two squares?
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B: \(x^2 + 16\)
A difference of two squares needs a minus sign. \(x^2 + 16\) is a sum of squares, so it does not factorise this way.
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9
Using \(a^2 - b^2 = (a + b)(a - b)\), work out \(52^2 - 48^2\).
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B: 400
\((52 + 48)(52 - 48) = 100 \times 4 = 400\).
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10
Factorise \(2x^2 + 7x + 3\).
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B: \((2x + 1)(x + 3)\)
\(2 \times 3 = 6\), and 6 and 1 add to 7, so \(2x^2 + 6x + x + 3 = 2x(x + 3) + (x + 3) = (2x + 1)(x + 3)\).