Revision guide by The Full Marks
Algebraic fractions: what you need to know Algebraic fractions follow exactly the same rules as ordinary fractions, but the numerators and denominators are expressions. This Higher-only topic appears near the end of the paper and relies on good factorising.
Key facts You can only cancel factors, never single terms: factorise the top and bottom first. Add or subtract with a common denominator: a b + c d = a d + b c b d \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd} b a + d c = b d a d + b c Multiply tops and bottoms; to divide, flip the second fraction and multiply. Look for the difference of two squares, e.g. x 2 − 9 = ( x + 3 ) ( x − 3 ) x^2 - 9 = (x + 3)(x - 3) x 2 − 9 = ( x + 3 ) ( x − 3 ) Write a whole number over the common denominator too, e.g. 2 = 2 ( x + 3 ) x + 3 2 = \frac{2(x + 3)}{x + 3} 2 = x + 3 2 ( x + 3 ) Simplifying Factorise the numerator and denominator fully (take out common factors first, and watch for non-unit quadratics such as 2 x 2 − 5 x − 3 = ( 2 x + 1 ) ( x − 3 ) 2x^2 - 5x - 3 = (2x + 1)(x - 3) 2 x 2 − 5 x − 3 = ( 2 x + 1 ) ( x − 3 ) Cancel any bracket that appears on both top and bottom. Adding or subtracting Use the product of the denominators (or their LCM) as the common denominator. Multiply each numerator by what its denominator was multiplied by, using brackets. Expand the top carefully, especially after a minus sign, and simplify. Solving an equation with algebraic fractions Multiply every term, including any number on the right-hand side, by the common denominator so all the fractions disappear. Expand the brackets and rearrange into the form a x 2 + b x + c = 0 ax^2 + bx + c = 0 a x 2 + b x + c = 0 Solve by factorising or the quadratic formula, giving surds in simplest form if asked. Worked examples Example 1 Higher · grade 7
Simplify fully x 2 − 9 x 2 + 5 x + 6 \frac{x^2 - 9}{x^2 + 5x + 6} x 2 + 5 x + 6 x 2 − 9
Show the solution Top: x 2 − 9 = ( x + 3 ) ( x − 3 ) x^2 - 9 = (x + 3)(x - 3) x 2 − 9 = ( x + 3 ) ( x − 3 ) x 2 + 5 x + 6 = ( x + 2 ) ( x + 3 ) x^2 + 5x + 6 = (x + 2)(x + 3) x 2 + 5 x + 6 = ( x + 2 ) ( x + 3 ) ( x + 3 ) (x + 3) ( x + 3 ) x − 3 x + 2 \frac{x - 3}{x + 2} x + 2 x − 3
Example 2 Higher · grade 8
Write 3 x + 2 − 2 x − 1 \frac{3}{x + 2} - \frac{2}{x - 1} x + 2 3 − x − 1 2
Show the solution Common denominator: ( x + 2 ) ( x − 1 ) (x + 2)(x - 1) ( x + 2 ) ( x − 1 ) 3 ( x − 1 ) − 2 ( x + 2 ) ( x + 2 ) ( x − 1 ) \frac{3(x - 1) - 2(x + 2)}{(x + 2)(x - 1)} ( x + 2 ) ( x − 1 ) 3 ( x − 1 ) − 2 ( x + 2 ) 3 x − 3 − 2 x − 4 = x − 7 3x - 3 - 2x - 4 = x - 7 3 x − 3 − 2 x − 4 = x − 7 x − 7 ( x + 2 ) ( x − 1 ) \frac{x - 7}{(x + 2)(x - 1)} ( x + 2 ) ( x − 1 ) x − 7
Example 3 Higher · grade 8
Solve 3 x + 1 + 2 x = 1 \frac{3}{x + 1} + \frac{2}{x} = 1 x + 1 3 + x 2 = 1 a ± b a \pm \sqrt{b} a ± b
Show the solution Multiply every term by x ( x + 1 ) x(x + 1) x ( x + 1 ) 3 x + 2 ( x + 1 ) = x ( x + 1 ) 3x + 2(x + 1) = x(x + 1) 3 x + 2 ( x + 1 ) = x ( x + 1 ) 5 x + 2 = x 2 + x 5x + 2 = x^2 + x 5 x + 2 = x 2 + x x 2 − 4 x − 2 = 0 x^2 - 4x - 2 = 0 x 2 − 4 x − 2 = 0 x = 4 ± 16 + 8 2 = 4 ± 2 6 2 x = \frac{4 \pm \sqrt{16 + 8}}{2} = \frac{4 \pm 2\sqrt{6}}{2} x = 2 4 ± 16 + 8 = 2 4 ± 2 6 x = 2 ± 6 x = 2 \pm \sqrt{6} x = 2 ± 6
Common mistakes Cancelling terms rather than factors, such as crossing out x 2 x^2 x 2 x 2 − 9 x 2 + 5 x + 6 \frac{x^2 - 9}{x^2 + 5x + 6} x 2 + 5 x + 6 x 2 − 9 Adding tops and bottoms separately: 3 x + 4 x + 1 \frac{3}{x} + \frac{4}{x + 1} x 3 + x + 1 4 7 2 x + 1 \frac{7}{2x + 1} 2 x + 1 7 Not multiplying the minus sign through the second bracket when subtracting. Forgetting to flip the second fraction when dividing. When solving, forgetting to multiply the number on the right-hand side by the common denominator. Exam tips There are usually marks for correct factorising and for the common denominator, even if the final answer slips, so show each line. In a "show that" question, your last line must match the given form exactly. Questions people ask How do you simplify algebraic fractions? Factorise the numerator and the denominator completely, then cancel any brackets that are the same on the top and bottom.
How do you add algebraic fractions? Find a common denominator, usually by multiplying the two denominators, adjust each numerator to match, then combine and simplify the top.
Exam questions: Algebraic fractions (19, easiest first) © Pearson Education Ltd. November 2024 Paper 2 (Higher, calculator), question 17.
Answer 2 y \frac{2}{y} y 2 k = 2 k = 2 k = 2 3 marks
Worked solution Checked against the mark scheme
Common denominator 10 x y 10xy 10 x y
6 x − y 10 x y + 5 y 10 x y − 2 ( 2 y − 7 x ) 10 x y \frac{6x - y}{10xy} + \frac{5y}{10xy} - \frac{2(2y - 7x)}{10xy} 10 x y 6 x − y + 10 x y 5 y − 10 x y 2 ( 2 y − 7 x )
= 6 x − y + 5 y − 4 y + 14 x 10 x y = 20 x 10 x y = 2 y = \frac{6x - y + 5y - 4y + 14x}{10xy} = \frac{20x}{10xy} = \frac{2}{y} = 10 x y 6 x − y + 5 y − 4 y + 14 x = 10 x y 20 x = y 2
Watch out: Not multiplying the minus sign through the bracket, getting − 14 x -14x − 14 x + 14 x +14x + 14 x
© Pearson Education Ltd. June 2024 Paper 3 (Higher, calculator), question 15.
(a) a − 3 5 \dfrac{a-3}{5} 5 a − 3 1 mark (b) ( 3 k − 1 ) ( k + 4 ) (3k-1)(k+4) ( 3 k − 1 ) ( k + 4 ) 2 marks (c) 2 − x x \dfrac{2-x}{x} x 2 − x 3 marks
Worked solution Checked against the mark scheme
(a) Cancel one factor of ( a − 3 ) (a-3) ( a − 3 ) a − 3 5 \dfrac{a-3}{5} 5 a − 3
(b) 3 k 2 + 11 k − 4 = ( 3 k − 1 ) ( k + 4 ) 3k^2 + 11k - 4 = (3k - 1)(k + 4) 3 k 2 + 11 k − 4 = ( 3 k − 1 ) ( k + 4 ) 3 k 2 + 12 k − k − 4 3k^2 + 12k - k - 4 3 k 2 + 12 k − k − 4
(c) ( 2 − x ) ( 2 + x ) x ( x + 3 ) × x + 3 x + 2 = 2 − x x \dfrac{(2-x)(2+x)}{x(x+3)} \times \dfrac{x+3}{x+2} = \dfrac{2-x}{x} x ( x + 3 ) ( 2 − x ) ( 2 + x ) × x + 2 x + 3 = x 2 − x
Watch out: In (c), not recognising 4 − x 2 4 - x^2 4 − x 2
© Pearson Education Ltd. June 2023 Paper 2 (Higher, calculator), question 14.
Answer x + 2 2 x + 1 \frac{x + 2}{2x + 1} 2 x + 1 x + 2 3 marks
Worked solution Checked against the mark scheme
Top: x 2 − x − 6 = ( x − 3 ) ( x + 2 ) x^2 - x - 6 = (x - 3)(x + 2) x 2 − x − 6 = ( x − 3 ) ( x + 2 )
Bottom: 2 x 2 − 5 x − 3 = ( x − 3 ) ( 2 x + 1 ) 2x^2 - 5x - 3 = (x - 3)(2x + 1) 2 x 2 − 5 x − 3 = ( x − 3 ) ( 2 x + 1 )
Cancel ( x − 3 ) (x - 3) ( x − 3 ) x + 2 2 x + 1 \frac{x + 2}{2x + 1} 2 x + 1 x + 2 a = 1 , b = 2 , c = 2 , d = 1 a = 1, b = 2, c = 2, d = 1 a = 1 , b = 2 , c = 2 , d = 1
Watch out: Factorising 2 x 2 − 5 x − 3 2x^2 - 5x - 3 2 x 2 − 5 x − 3 ( 2 x − 3 ) ( x + 1 ) (2x - 3)(x + 1) ( 2 x − 3 ) ( x + 1 )
© Pearson Education Ltd. November 2021 Paper 1 (Higher, non-calculator), question 15.
Answer 26 x + 15 10 x \frac{26x + 15}{10x} 10 x 26 x + 15 3 marks
Worked solution Checked against the mark scheme
Common denominator 10 x 10x 10 x
4 x + 3 2 x = 5 ( 4 x + 3 ) 10 x = 20 x + 15 10 x \frac{4x + 3}{2x} = \frac{5(4x + 3)}{10x} = \frac{20x + 15}{10x} 2 x 4 x + 3 = 10 x 5 ( 4 x + 3 ) = 10 x 20 x + 15 3 5 = 6 x 10 x \frac{3}{5} = \frac{6x}{10x} 5 3 = 10 x 6 x
Add: 20 x + 15 + 6 x 10 x = 26 x + 15 10 x \frac{20x + 15 + 6x}{10x} = \frac{26x + 15}{10x} 10 x 20 x + 15 + 6 x = 10 x 26 x + 15
So a = 26 a = 26 a = 26 b = 15 b = 15 b = 15 c = 10 c = 10 c = 10
Watch out: Adding numerators and denominators directly to get 4 x + 6 2 x + 5 \frac{4x + 6}{2x + 5} 2 x + 5 4 x + 6
© Pearson Education Ltd. November 2020 Paper 3 (Higher, calculator), question 12.
(a) 3 x 2 ( x + 2 ) ( x − 4 ) \frac{3x^2}{(x+2)(x-4)} ( x + 2 ) ( x − 4 ) 3 x 2 3 marks (b) 8 x 3 − 2 x 2 − 51 x − 45 8x^3 - 2x^2 - 51x - 45 8 x 3 − 2 x 2 − 51 x − 45 3 marks
Worked solution Checked against the mark scheme
(a) x ( x − 4 ) + 2 x ( x + 2 ) ( x + 2 ) ( x − 4 ) = x 2 − 4 x + 2 x 2 + 4 x ( x + 2 ) ( x − 4 ) = 3 x 2 ( x + 2 ) ( x − 4 ) \frac{x(x-4) + 2x(x+2)}{(x+2)(x-4)} = \frac{x^2 - 4x + 2x^2 + 4x}{(x+2)(x-4)} = \frac{3x^2}{(x+2)(x-4)} ( x + 2 ) ( x − 4 ) x ( x − 4 ) + 2 x ( x + 2 ) = ( x + 2 ) ( x − 4 ) x 2 − 4 x + 2 x 2 + 4 x = ( x + 2 ) ( x − 4 ) 3 x 2
(b) ( x − 3 ) ( 2 x + 3 ) = 2 x 2 − 3 x − 9 (x-3)(2x+3) = 2x^2 - 3x - 9 ( x − 3 ) ( 2 x + 3 ) = 2 x 2 − 3 x − 9
( 2 x 2 − 3 x − 9 ) ( 4 x + 5 ) = 8 x 3 + 10 x 2 − 12 x 2 − 15 x − 36 x − 45 (2x^2 - 3x - 9)(4x + 5) = 8x^3 + 10x^2 - 12x^2 - 15x - 36x - 45 ( 2 x 2 − 3 x − 9 ) ( 4 x + 5 ) = 8 x 3 + 10 x 2 − 12 x 2 − 15 x − 36 x − 45
= 8 x 3 − 2 x 2 − 51 x − 45 = 8x^3 - 2x^2 - 51x - 45 = 8 x 3 − 2 x 2 − 51 x − 45
Watch out: Adding the numerators and denominators separately in (a).
© Pearson Education Ltd. November 2019 Paper 1 (Higher, non-calculator), question 13.
(a) 17 x + 2 3 x ( x + 1 ) \frac{17x+2}{3x(x+1)} 3 x ( x + 1 ) 17 x + 2 2 marks (b) ( x + y ) ( x + y + 3 ) (x+y)(x+y+3) ( x + y ) ( x + y + 3 ) 1 mark
Worked solution Checked against the mark scheme
(a) Common denominator 3 x ( x + 1 ) 3x(x+1) 3 x ( x + 1 )
5 × 3 x 3 x ( x + 1 ) + 2 ( x + 1 ) 3 x ( x + 1 ) = 15 x + 2 x + 2 3 x ( x + 1 ) = 17 x + 2 3 x ( x + 1 ) \frac{5 \times 3x}{3x(x+1)} + \frac{2(x+1)}{3x(x+1)} = \frac{15x + 2x + 2}{3x(x+1)} = \frac{17x+2}{3x(x+1)} 3 x ( x + 1 ) 5 × 3 x + 3 x ( x + 1 ) 2 ( x + 1 ) = 3 x ( x + 1 ) 15 x + 2 x + 2 = 3 x ( x + 1 ) 17 x + 2
(b) ( x + y ) (x+y) ( x + y )
( x + y ) 2 + 3 ( x + y ) = ( x + y ) ( x + y + 3 ) (x+y)^2 + 3(x+y) = (x+y)(x+y+3) ( x + y ) 2 + 3 ( x + y ) = ( x + y ) ( x + y + 3 )
Watch out: In (a), adding numerators and denominators to get 7/(4x+1).
© Pearson Education Ltd. June 2019 Paper 2 (Higher, calculator), question 13.
Answer 7 x − 13 x − 2 \frac{7x - 13}{x - 2} x − 2 7 x − 13 4 marks
Worked solution Checked against the mark scheme
x 2 + 3 x − 10 = ( x + 5 ) ( x − 2 ) x^2 + 3x - 10 = (x + 5)(x - 2) x 2 + 3 x − 10 = ( x + 5 ) ( x − 2 )
( x + 5 ) ÷ ( x + 5 ) ( x − 2 ) x − 1 = ( x + 5 ) × x − 1 ( x + 5 ) ( x − 2 ) = x − 1 x − 2 (x + 5) \div \frac{(x + 5)(x - 2)}{x - 1} = (x + 5) \times \frac{x - 1}{(x + 5)(x - 2)} = \frac{x - 1}{x - 2} ( x + 5 ) ÷ x − 1 ( x + 5 ) ( x − 2 ) = ( x + 5 ) × ( x + 5 ) ( x − 2 ) x − 1 = x − 2 x − 1
6 + x − 1 x − 2 = 6 ( x − 2 ) + x − 1 x − 2 = 7 x − 13 x − 2 6 + \frac{x - 1}{x - 2} = \frac{6(x - 2) + x - 1}{x - 2} = \frac{7x - 13}{x - 2} 6 + x − 2 x − 1 = x − 2 6 ( x − 2 ) + x − 1 = x − 2 7 x − 13
Watch out: Multiplying by the fraction instead of by its reciprocal.
© Pearson Education Ltd. June 2018 Paper 1 (Higher, non-calculator), question 17.
Answer 3 x + 1 2 x \frac{3x + 1}{2x} 2 x 3 x + 1 3 marks
Worked solution Checked against the mark scheme
Numerator: 3 x 2 − 8 x − 3 = ( 3 x + 1 ) ( x − 3 ) 3x^2 - 8x - 3 = (3x + 1)(x - 3) 3 x 2 − 8 x − 3 = ( 3 x + 1 ) ( x − 3 )
Denominator: 2 x 2 − 6 x = 2 x ( x − 3 ) 2x^2 - 6x = 2x(x - 3) 2 x 2 − 6 x = 2 x ( x − 3 )
Cancel ( x − 3 ) (x - 3) ( x − 3 ) 3 x + 1 2 x \frac{3x + 1}{2x} 2 x 3 x + 1
Watch out: Cancelling single terms such as the x 2 x^2 x 2
© Pearson Education Ltd. June 2023 Paper 3 (Higher, calculator), question 21.
Answer x = − 2 x = -2 x = − 2 x = − 3 2 x = -\frac{3}{2} x = − 2 3 4 marks
Worked solution Checked against the mark scheme
Multiply every term by ( x + 4 ) ( 2 − 2 x ) (x + 4)(2 - 2x) ( x + 4 ) ( 2 − 2 x )
( 2 − 2 x ) + 3 ( x + 4 ) = ( x + 4 ) ( 2 − 2 x ) (2 - 2x) + 3(x + 4) = (x + 4)(2 - 2x) ( 2 − 2 x ) + 3 ( x + 4 ) = ( x + 4 ) ( 2 − 2 x )
x + 14 = − 2 x 2 − 6 x + 8 x + 14 = -2x^2 - 6x + 8 x + 14 = − 2 x 2 − 6 x + 8
2 x 2 + 7 x + 6 = 0 2x^2 + 7x + 6 = 0 2 x 2 + 7 x + 6 = 0
( 2 x + 3 ) ( x + 2 ) = 0 (2x + 3)(x + 2) = 0 ( 2 x + 3 ) ( x + 2 ) = 0 x = − 3 2 x = -\frac{3}{2} x = − 2 3 x = − 2 x = -2 x = − 2
Watch out: Forgetting to multiply the right-hand side 1 by both denominators.
© Pearson Education Ltd. June 2022 Paper 1 (Higher, non-calculator), question 19.
Answer x = 5 ± 15 2 x = \frac{5 \pm \sqrt{15}}{2} x = 2 5 ± 15 4 marks
Worked solution Checked against the mark scheme
Multiply by ( 2 x − 1 ) ( x − 1 ) (2x-1)(x-1) ( 2 x − 1 ) ( x − 1 ) ( x − 1 ) + 3 ( 2 x − 1 ) = ( 2 x − 1 ) ( x − 1 ) (x - 1) + 3(2x - 1) = (2x - 1)(x - 1) ( x − 1 ) + 3 ( 2 x − 1 ) = ( 2 x − 1 ) ( x − 1 )
7 x − 4 = 2 x 2 − 3 x + 1 7x - 4 = 2x^2 - 3x + 1 7 x − 4 = 2 x 2 − 3 x + 1 2 x 2 − 10 x + 5 = 0 2x^2 - 10x + 5 = 0 2 x 2 − 10 x + 5 = 0
x = 10 ± 100 − 40 4 = 10 ± 2 15 4 = 5 ± 15 2 x = \frac{10 \pm \sqrt{100 - 40}}{4} = \frac{10 \pm 2\sqrt{15}}{4} = \frac{5 \pm \sqrt{15}}{2} x = 4 10 ± 100 − 40 = 4 10 ± 2 15 = 2 5 ± 15
Watch out: Not multiplying the 1 on the right-hand side by the common denominator.
© Pearson Education Ltd. November 2022 Paper 1 (Higher, non-calculator), question 19.
Answer x = − 1 2 ± 1 2 2 x = -\frac{1}{2} \pm \frac{1}{2}\sqrt{2} x = − 2 1 ± 2 1 2 5 marks
Worked solution Checked against the mark scheme
Combine the fractions: ( x + 1 ) − x x ( x + 1 ) = 4 \frac{(x + 1) - x}{x(x + 1)} = 4 x ( x + 1 ) ( x + 1 ) − x = 4 1 x ( x + 1 ) = 4 \frac{1}{x(x + 1)} = 4 x ( x + 1 ) 1 = 4
1 = 4 x 2 + 4 x 1 = 4x^2 + 4x 1 = 4 x 2 + 4 x 4 x 2 + 4 x − 1 = 0 4x^2 + 4x - 1 = 0 4 x 2 + 4 x − 1 = 0
x = − 4 ± 16 + 16 8 = − 4 ± 4 2 8 x = \frac{-4 \pm \sqrt{16 + 16}}{8} = \frac{-4 \pm 4\sqrt{2}}{8} x = 8 − 4 ± 16 + 16 = 8 − 4 ± 4 2
x = − 1 2 ± 1 2 2 x = -\frac{1}{2} \pm \frac{1}{2}\sqrt{2} x = − 2 1 ± 2 1 2
Watch out: Writing 32 \sqrt{32} 32 16 2 16\sqrt{2} 16 2 4 2 4\sqrt{2} 4 2
© Pearson Education Ltd. June 2022 Paper 3 (Higher, calculator), question 19.
Answer − 11 x + 2 x 2 − 4 \frac{-11x + 2}{x^2 - 4} x 2 − 4 − 11 x + 2 a = − 11 a = -11 a = − 11 b = 2 b = 2 b = 2 4 marks
Worked solution Checked against the mark scheme
Common denominator ( x + 2 ) ( x − 2 ) = x 2 − 4 (x + 2)(x - 2) = x^2 - 4 ( x + 2 ) ( x − 2 ) = x 2 − 4
3 x ( x − 2 ) − ( 2 x + 1 ) ( x + 2 ) − ( x 2 − 4 ) x 2 − 4 \frac{3x(x - 2) - (2x + 1)(x + 2) - (x^2 - 4)}{x^2 - 4} x 2 − 4 3 x ( x − 2 ) − ( 2 x + 1 ) ( x + 2 ) − ( x 2 − 4 )
Numerator: 3 x 2 − 6 x − ( 2 x 2 + 5 x + 2 ) − x 2 + 4 = − 11 x + 2 3x^2 - 6x - (2x^2 + 5x + 2) - x^2 + 4 = -11x + 2 3 x 2 − 6 x − ( 2 x 2 + 5 x + 2 ) − x 2 + 4 = − 11 x + 2
So the expression is − 11 x + 2 x 2 − 4 \frac{-11x + 2}{x^2 - 4} x 2 − 4 − 11 x + 2
Watch out: Not distributing the minus sign across ( 2 x + 1 ) ( x + 2 ) (2x + 1)(x + 2) ( 2 x + 1 ) ( x + 2 )
© Pearson Education Ltd. November 2022 Paper 3 (Higher, calculator), question 22.
Answer a = 4 a = 4 a = 4 b = 110 b = 110 b = 110 3 marks
Worked solution Checked against the mark scheme
Common denominator: ( x − 5 ) ( x + 5 ) = x 2 − 25 (x - 5)(x + 5) = x^2 - 25 ( x − 5 ) ( x + 5 ) = x 2 − 25
Numerator: ( 2 x + 3 ) ( x + 5 ) + ( x − 4 ) ( x − 5 ) − 3 ( x 2 − 25 ) (2x + 3)(x + 5) + (x - 4)(x - 5) - 3(x^2 - 25) ( 2 x + 3 ) ( x + 5 ) + ( x − 4 ) ( x − 5 ) − 3 ( x 2 − 25 )
= 2 x 2 + 13 x + 15 + x 2 − 9 x + 20 − 3 x 2 + 75 = 4 x + 110 = 2x^2 + 13x + 15 + x^2 - 9x + 20 - 3x^2 + 75 = 4x + 110 = 2 x 2 + 13 x + 15 + x 2 − 9 x + 20 − 3 x 2 + 75 = 4 x + 110
So a = 4 a = 4 a = 4 b = 110 b = 110 b = 110
Watch out: Writing − 3 -3 − 3 − 3 -3 − 3 x 2 − 25 x^2 - 25 x 2 − 25
© Pearson Education Ltd. November 2021 Paper 3 (Higher, calculator), question 22.
Answer 1 x ( x + 4 ) \frac{1}{x(x + 4)} x ( x + 4 ) 1 r = 4 r = 4 r = 4 3 marks
Worked solution Checked against the mark scheme
6 x 3 9 x 2 − 144 × 3 ( x − 4 ) 2 x 4 \frac{6x^3}{9x^2 - 144} \times \frac{3(x - 4)}{2x^4} 9 x 2 − 144 6 x 3 × 2 x 4 3 ( x − 4 )
9 x 2 − 144 = 9 ( x − 4 ) ( x + 4 ) 9x^2 - 144 = 9(x - 4)(x + 4) 9 x 2 − 144 = 9 ( x − 4 ) ( x + 4 )
= 18 x 3 ( x − 4 ) 18 x 4 ( x − 4 ) ( x + 4 ) = 1 x ( x + 4 ) = \frac{18x^3(x - 4)}{18x^4(x - 4)(x + 4)} = \frac{1}{x(x + 4)} = 18 x 4 ( x − 4 ) ( x + 4 ) 18 x 3 ( x − 4 ) = x ( x + 4 ) 1 r = 4 r = 4 r = 4
Watch out: Factorising 9 x 2 − 144 9x^2 - 144 9 x 2 − 144 ( 3 x − 12 ) 2 (3x - 12)^2 ( 3 x − 12 ) 2
© Pearson Education Ltd. November 2019 Paper 3 (Higher, calculator), question 22.
Worked solution Checked against the mark scheme
7 x − 14 = 7 ( x − 2 ) 7x - 14 = 7(x - 2) 7 x − 14 = 7 ( x − 2 ) x 2 + 4 x − 12 = ( x − 2 ) ( x + 6 ) x^2 + 4x - 12 = (x - 2)(x + 6) x 2 + 4 x − 12 = ( x − 2 ) ( x + 6 )
x 3 − 36 x = x ( x − 6 ) ( x + 6 ) x^3 - 36x = x(x - 6)(x + 6) x 3 − 36 x = x ( x − 6 ) ( x + 6 )
7 ( x − 2 ) ( x − 2 ) ( x + 6 ) × x ( x − 6 ) ( x + 6 ) x − 6 \dfrac{7(x-2)}{(x-2)(x+6)} \times \dfrac{x(x-6)(x+6)}{x-6} ( x − 2 ) ( x + 6 ) 7 ( x − 2 ) × x − 6 x ( x − 6 ) ( x + 6 )
Cancel ( x − 2 ) (x - 2) ( x − 2 ) ( x + 6 ) (x + 6) ( x + 6 ) ( x − 6 ) (x - 6) ( x − 6 ) 7 x 7x 7 x a = 7 a = 7 a = 7
Watch out: Factorising x 3 − 36 x x^3 - 36x x 3 − 36 x x ( x − 6 ) 2 x(x - 6)^2 x ( x − 6 ) 2 x ( x 2 − 36 ) x(x^2 - 36) x ( x 2 − 36 )
© Pearson Education Ltd. November 2018 Paper 2 (Higher, calculator), question 12.
(a) 4 x − 6 3 x − 9 \frac{4x - 6}{3x - 9} 3 x − 9 4 x − 6 3 marks (b) 8 − x x ( x + 1 ) ( x − 2 ) \frac{8 - x}{x(x+1)(x-2)} x ( x + 1 ) ( x − 2 ) 8 − x 3 marks
Worked solution Checked against the mark scheme
(a) ( 2 x − 3 ) ( 2 x + 3 ) 3 ( 2 x + 3 ) × 2 x x ( x − 3 ) = 2 ( 2 x − 3 ) 3 ( x − 3 ) = 4 x − 6 3 x − 9 \frac{(2x-3)(2x+3)}{3(2x+3)} \times \frac{2x}{x(x-3)} = \frac{2(2x-3)}{3(x-3)} = \frac{4x-6}{3x-9} 3 ( 2 x + 3 ) ( 2 x − 3 ) ( 2 x + 3 ) × x ( x − 3 ) 2 x = 3 ( x − 3 ) 2 ( 2 x − 3 ) = 3 x − 9 4 x − 6
(b) Common denominator x ( x + 1 ) ( x − 2 ) x(x+1)(x-2) x ( x + 1 ) ( x − 2 )
Numerator: 3 x ( x − 2 ) + x ( x + 1 ) − 4 ( x + 1 ) ( x − 2 ) 3x(x-2) + x(x+1) - 4(x+1)(x-2) 3 x ( x − 2 ) + x ( x + 1 ) − 4 ( x + 1 ) ( x − 2 ) = 3 x 2 − 6 x + x 2 + x − 4 x 2 + 4 x + 8 = 8 − x = 3x^2 - 6x + x^2 + x - 4x^2 + 4x + 8 = 8 - x = 3 x 2 − 6 x + x 2 + x − 4 x 2 + 4 x + 8 = 8 − x
Answer: 8 − x x ( x + 1 ) ( x − 2 ) \frac{8 - x}{x(x+1)(x-2)} x ( x + 1 ) ( x − 2 ) 8 − x
Watch out: Sign errors when subtracting 4 ( x + 1 ) ( x − 2 ) 4(x+1)(x-2) 4 ( x + 1 ) ( x − 2 )
© Pearson Education Ltd. Specimen Paper 1 (Higher, non-calculator), question 15.
Answer 2 x − 5 x + 5 \frac{2x - 5}{x + 5} x + 5 2 x − 5 3 marks
Worked solution Checked against the mark scheme
Numerator: 2 x 2 − 3 x − 5 = ( 2 x − 5 ) ( x + 1 ) 2x^2 - 3x - 5 = (2x - 5)(x + 1) 2 x 2 − 3 x − 5 = ( 2 x − 5 ) ( x + 1 )
Denominator: x 2 + 6 x + 5 = ( x + 5 ) ( x + 1 ) x^2 + 6x + 5 = (x + 5)(x + 1) x 2 + 6 x + 5 = ( x + 5 ) ( x + 1 )
Cancel the common factor ( x + 1 ) (x + 1) ( x + 1 )
2 x − 5 x + 5 \frac{2x - 5}{x + 5} x + 5 2 x − 5 a = 2 a = 2 a = 2 b = − 5 b = -5 b = − 5 c = 1 c = 1 c = 1 d = 5 d = 5 d = 5
Watch out: Cancelling the x 2 x^2 x 2
© Pearson Education Ltd. Specimen Paper 3 (Higher, calculator), question 14.
Answer 3 x + 10 x + 2 \frac{3x + 10}{x + 2} x + 2 3 x + 10 4 marks
Worked solution Checked against the mark scheme
x 2 + 5 x + 6 = ( x + 2 ) ( x + 3 ) x^2 + 5x + 6 = (x + 2)(x + 3) x 2 + 5 x + 6 = ( x + 2 ) ( x + 3 )
( x + 3 ) ÷ ( x + 2 ) ( x + 3 ) x − 2 = ( x + 3 ) × x − 2 ( x + 2 ) ( x + 3 ) = x − 2 x + 2 (x + 3) \div \frac{(x + 2)(x + 3)}{x - 2} = (x + 3) \times \frac{x - 2}{(x + 2)(x + 3)} = \frac{x - 2}{x + 2} ( x + 3 ) ÷ x − 2 ( x + 2 ) ( x + 3 ) = ( x + 3 ) × ( x + 2 ) ( x + 3 ) x − 2 = x + 2 x − 2
4 − x − 2 x + 2 = 4 ( x + 2 ) − ( x − 2 ) x + 2 = 4 x + 8 − x + 2 x + 2 = 3 x + 10 x + 2 4 - \frac{x - 2}{x + 2} = \frac{4(x + 2) - (x - 2)}{x + 2} = \frac{4x + 8 - x + 2}{x + 2} = \frac{3x + 10}{x + 2} 4 − x + 2 x − 2 = x + 2 4 ( x + 2 ) − ( x − 2 ) = x + 2 4 x + 8 − x + 2 = x + 2 3 x + 10
Watch out: Subtracting only the x from the numerator, giving 3x + 6 instead of 3x + 10.
© Pearson Education Ltd. November 2023 Paper 2 (Higher, calculator), question 20.
Answer 6 x + 37 6 x − 42 \frac{6x + 37}{6x - 42} 6 x − 42 6 x + 37 4 marks
Worked solution Checked against the mark scheme
2 x 2 − 6 x − 56 = 2 ( x + 4 ) ( x − 7 ) 2x^2 - 6x - 56 = 2(x + 4)(x - 7) 2 x 2 − 6 x − 56 = 2 ( x + 4 ) ( x − 7 ) 3 x − 21 = 3 ( x − 7 ) 3x - 21 = 3(x - 7) 3 x − 21 = 3 ( x − 7 )
( x + 4 ) × 2 x + 3 2 ( x + 4 ) ( x − 7 ) = 2 x + 3 2 ( x − 7 ) (x + 4) \times \frac{2x + 3}{2(x + 4)(x - 7)} = \frac{2x + 3}{2(x - 7)} ( x + 4 ) × 2 ( x + 4 ) ( x − 7 ) 2 x + 3 = 2 ( x − 7 ) 2 x + 3
14 3 ( x − 7 ) + 2 x + 3 2 ( x − 7 ) = 28 + 3 ( 2 x + 3 ) 6 ( x − 7 ) = 6 x + 37 6 x − 42 \frac{14}{3(x - 7)} + \frac{2x + 3}{2(x - 7)} = \frac{28 + 3(2x + 3)}{6(x - 7)} = \frac{6x + 37}{6x - 42} 3 ( x − 7 ) 14 + 2 ( x − 7 ) 2 x + 3 = 6 ( x − 7 ) 28 + 3 ( 2 x + 3 ) = 6 x − 42 6 x + 37
Watch out: Multiplying by the divisor instead of its reciprocal.
Also tested in these questions Their main topic is different, but part of each one tests this topic.
November 2024 · Paper 1 Higher · Q11 : Correct a mix-up about the sum and product when factorising a quadratic, then factorise a difference of two squares and an expression by grouping.November 2024 · Paper 1 Higher · Q13 : Form and solve an equation from the difference in volume of two cuboids with algebraic side lengths.November 2024 · Paper 2 Foundation · Q11 : Simplify a product of two terms and collect like terms in an expression.June 2024 · Paper 1 Foundation · Q17 : Solve a linear equation with brackets.June 2024 · Paper 1 Higher · Q20 : Use index laws with a cube root and a power of root 2 to find an unknown power.November 2023 · Paper 3 Higher · Q23 : Solve an equation of algebraic fractions in x squared and y squared to find the ratio x to y in its simplest form.June 2023 · Paper 2 Foundation · Q4 : Simplify a product of a number and an algebraic term.June 2023 · Paper 3 Higher · Q24 : Find an algebraic probability of picking two counters of the same colour without replacement, using a ratio to write everything in one variable.November 2022 · Paper 1 Higher · Q16 : Prove an algebraic identity for the difference of two squared expressions, then judge a claim about odd numbers that follows from it.November 2022 · Paper 1 Foundation · Q21 : Simplify a squared product of powers of 2 and give the answer as a single power.November 2022 · Paper 1 Foundation · Q26 : Solve a linear inequality with a fraction, then factorise a quadratic.June 2022 · Paper 1 Foundation · Q2 : Collect four identical letter terms into a single term.June 2022 · Paper 3 Foundation · Q28 : Show that the area of a shaded trapezium in a rectangle is a given quadratic expression in x.November 2021 · Paper 3 Foundation · Q6 : Simplify a product of letters and a number, then collect like terms in an expression.November 2020 · Paper 1 Foundation · Q30 : Rearrange a linear formula to change the subject, and simplify a negative power raised to a negative power.November 2020 · Paper 2 Higher · Q10 : Simplify an expression raised to the power 0, cancel an algebraic fraction, and raise a product of powers to a power.November 2020 · Paper 2 Higher · Q14 : Set up and solve an equation for two squares whose sides differ by 4 cm and areas differ by 70 cm², then find the larger area.November 2019 · Paper 1 Foundation · Q13 : Simplify a product of two terms, then collect like terms in an expression.November 2019 · Paper 3 Higher · Q10 : Solve a linear equation with a fraction, and simplify an algebraic fraction with a repeated bracket.June 2019 · Paper 1 Higher · Q21 : Find an inverse function, then use composite functions to show that a given quadratic equation holds.June 2019 · Paper 2 Foundation · Q8 : Simplify three algebraic products, the last one a fraction of letters that cancels.November 2018 · Paper 2 Foundation · Q19 : Solve a linear equation with a bracket and factorise an expression fully.June 2018 · Paper 1 Foundation · Q16 : Substitute values into a formula, expand a single bracket and solve a linear equation with brackets.Specimen · Paper 1 Higher · Q14 : Solve an equation with two algebraic fractions that have x in the denominators.Specimen · Paper 1 Higher · Q21 : Find an algebraic expression for the probability of picking one pen of each colour without replacement.Specimen · Paper 2 Foundation · Q4 : Simplify three short algebraic expressions by collecting like terms and multiplying.Specimen · Paper 2 Foundation · Q16 : Solve two linear equations and simplify a power raised to a power.Specimen · Paper 2 Higher · Q18 : Substitute x + 2 into a quadratic function and simplify to the form ax² + bx.Specimen · Paper 2 Higher · Q19 : Form and solve a quadratic from the area of a right-angled triangle, then use Pythagoras to find its perimeter.Specimen · Paper 3 Foundation · Q7 : Solve two simple linear equations and simplify a product of powers of d.
