Practise 29 real questions on Algebra from 2021 to 2026. See the answer and explanation as soon as you answer.
Algebra is one of the 8 main topics in our A Level Mathematics (Cambridge (CAIE) 9709) question bank, and about 6% of its real questions come from it. Tayyari Karo has 29 real questions on this topic from 2021 to 2026, from Paper 2, Paper 3. By difficulty: 19 medium, 7 hard, 3 easy.
Work through the questions below, then open the explanation for each one. Redo the ones you missed a few days later.
These are real past-paper questions. Think of your answer first, then open the explanation.
Solve the inequality 3|x + 4| > |2x − 1|. [4]
Square both sides: 9(x + 4)² > (2x − 1)², i.e. (3x + 12 − (2x − 1))(3x + 12 + (2x − 1)) > 0, so (x + 13)(5x + 11) > 0.
Critical values x = −13 and x = −11/5.
Solution: x < −13 or x > −11/5.
Square both sides (both are non-negative), use a difference of two squares to find the critical values, then choose the outer regions.
The polynomial p(x) is defined by p(x) = ax³ + bx² − ax − 24, where a and b are constants. It is given that (2x − 3) is a factor of p(x) and that the remainder is −15 when p(x) is divided by (x + 1).
(a) Find the values of a and b. [4]
(b) Hence factorise p(x) completely. [3]
(c) Hence solve the equation p(3cosecθ) = 0 for 90° < θ < 270°. [2]
(a) Since (2x−3) is a factor, p(3/2) = 0: 27a/8 + 9b/4 − 3a/2 − 24 = 0. Since the remainder on division by (x+1) is −15, p(−1) = −15: −a + b + a − 24 = −15, so b = 9. Substituting b = 9 into the first equation gives a = 2.
(b) With a = 2, b = 9: p(x) = 2x³ + 9x² − 2x − 24. Dividing by (2x − 3) gives quotient x² + 6x + 8 = (x+2)(x+4). So p(x) = (2x − 3)(x + 2)(x + 4).
(c) Setting p(3cosecθ) = 0: (2(3cosecθ) − 3)(3cosecθ + 2)(3cosecθ + 4) = 0, giving cosecθ = 1/2, −2/3, or −4/3, i.e. sinθ = 2, −3/2, or −3/4. Only sinθ = −3/4 is valid (|sinθ| ≤ 1). For 90° < θ < 270° with sinθ = −3/4 negative, θ = 228.6°.
Use the factor and remainder theorems to form simultaneous equations for a and b, factorise, then substitute 3cosecθ into the factorised form and discard invalid roots.
Expand (3 + x)(1 − 2x)^½ in ascending powers of x, up to and including the term in x², simplifying the coefficients. [4]
(1 − 2x)^½ = 1 + ½(−2x) + [½(−½)/2!](−2x)² + … = 1 − x − ½x² + …
(3 + x)(1 − x − ½x²) = 3(1 − x − ½x²) + x(1 − x − ½x²)
= (3 − 3x − (3/2)x²) + (x − x² + …)
= 3 − 2x − (5/2)x² (up to the term in x²).
Expand (1−2x)^½ by the binomial series, then multiply by (3+x) and collect terms up to x².
The polynomial 4x³ + ax² + 5x + b, where a and b are constants, is denoted by p(x). It is given that (2x + 1) is a factor of p(x). When p(x) is divided by (x − 4) the remainder is equal to 3 times the remainder when p(x) is divided by (x − 2).
Find the values of a and b. [5]
Since (2x+1) is a factor, p(−1/2) = 0:
4(−1/2)³ + a(−1/2)² + 5(−1/2) + b = 0
−1/2 + a/4 − 5/2 + b = 0 ⟹ a/4 + b = 3 ⟹ a + 4b = 12 … (1)
By the remainder theorem, the remainder on dividing by (x−4) is p(4), and by (x−2) is p(2), so p(4) = 3p(2):
p(4) = 4(64) + 16a + 20 + b = 276 + 16a + b
p(2) = 4(8) + 4a + 10 + b = 42 + 4a + b
276 + 16a + b = 3(42 + 4a + b) = 126 + 12a + 3b
150 + 4a − 2b = 0 ⟹ −2a + b = 75 … (2)
From (2): b = 75 + 2a. Substitute into (1): a + 4(75+2a) = 12 ⟹ 9a = −288 ⟹ a = −32
b = 75 + 2(−32) = 11
So a = −32 and b = 11.
Use the factor theorem to get one equation, then the remainder theorem twice more to get a second equation, and solve simultaneously.
The polynomial p(x) is defined by p(x) = 6x³ + ax² + bx − 20, where a and b are constants. It is given that x + 2 is a factor of p(x) and that the remainder is −11 when p(x) is divided by (x + 1).
(a) Find the values of a and b. [5]
(b) Hence factorise p(x), and determine the exact roots of the equation p(3x) = 0. [4]
(a) Since x + 2 is a factor, p(−2) = 0:
6(−8) + 4a − 2b − 20 = 0 ⟹ −48 + 4a − 2b − 20 = 0 ⟹ 4a − 2b − 68 = 0
Since the remainder on division by (x+1) is −11, p(−1) = −11:
6(−1) + a − b − 20 = −11 ⟹ −6 + a − b − 20 = −11 ⟹ a − b − 26 = −11 ⟹ a − b = 15
From the first equation: 2a − b = 34
Subtract: (2a − b) − (a − b) = 34 − 15 ⟹ a = 19
Then b = a − 15 = 4
So a = 19, b = 4.
(b) p(x) = 6x³ + 19x² + 4x − 20. Divide by (x + 2):
6x³ + 19x² + 4x − 20 = (x + 2)(6x² + 7x − 10)
Factorise the quadratic: 6x² + 7x − 10, discriminant = 49 + 240 = 289 = 17², roots = (−7 ± 17)/12 = 5/6 or −2
So 6x² + 7x − 10 = (6x − 5)(x + 2)
Thus p(x) = (x + 2)²(6x − 5). …
The full model answer is in the app.
(a) Expand (2 − x²)⁻² in ascending powers of x, up to and including the term in x⁴, simplifying the coefficients. [4]
(b) State the set of values of x for which the expansion is valid. [1]
(a) (2−x²)⁻² = 2⁻²(1 − x²/2)⁻² = (1/4)(1 − x²/2)⁻²
Using (1+u)ⁿ ≈ 1 + nu + [n(n−1)/2!]u² with n=−2, u=−x²/2:
(1 − x²/2)⁻² = 1 + (−2)(−x²/2) + [(−2)(−3)/2!](−x²/2)² + ...
= 1 + x² + 3(x⁴/4) + ...
= 1 + x² + (3/4)x⁴ + ...
Multiply by 1/4:
(2−x²)⁻² = 1/4 + (1/4)x² + (3/16)x⁴ + ...
(b) Valid when |x²/2| < 1, i.e. x² < 2, i.e. −√2 < x < √2
Take out a factor of 2⁻² to write the bracket as (1 − x²/2)⁻² and apply the binomial series for (1+u)ⁿ with n = −2.
Let f(x) = (2x² + 7x + 8)/[(1+x)(2+x)²].
(a) Express f(x) in partial fractions. [5]
(b) Hence obtain the expansion of f(x) in ascending powers of x, up to and including the term in x². [5]
(a) Write f(x) = A/(1+x) + B/(2+x) + C/(2+x)². Then 2x²+7x+8 = A(2+x)² + B(1+x)(2+x) + C(1+x). At x=-1: 3 = A. At x=-2: 2 = -C, so C=-2. Comparing x² coefficients: 2 = A+B, so B = -1. Hence f(x) = 3/(1+x) - 1/(2+x) - 2/(2+x)².
(b) Expanding each term: 3/(1+x) = 3(1-x+x²-...) = 3-3x+3x². -1/(2+x) = -½(1+x/2)⁻¹ = -½(1-x/2+x²/4-...) = -½+x/4-x²/8. -2/(2+x)² = -½(1+x/2)⁻² = -½(1-x+¾x²-...) = -½+x/2-3x²/8. Summing: constant term 3-½-½=2; x term -3+¼+½=-9/4; x² term 3-⅛-⅜=5/2. So f(x) = 2 - (9/4)x + (5/2)x² + ... .
Use the cover-up rule and coefficient comparison to find the partial fraction constants, then expand each fraction as a binomial series and collect terms.
The polynomial p(x) is defined by
p(x) = ax³ − 11x² − 19x − a,
where a is a constant. It is given that (x − 3) is a factor of p(x).
(a) Find the value of a. [2]
(b) When a has this value, factorise p(x) completely. [3]
(c) Hence find the exact values of y that satisfy the equation p(eʸ + e⁻ʸ) = 0. [4]
(a) Since (x − 3) is a factor, p(3) = 0:
a(27) − 11(9) − 19(3) − a = 0
27a − 99 − 57 − a = 0
26a − 156 = 0
a = 6
(b) With a = 6: p(x) = 6x³ − 11x² − 19x − 6
Dividing by (x − 3):
6x³ − 11x² − 19x − 6 = (x − 3)(6x² + 7x + 2)
Factorise the quadratic: 6x² + 7x + 2 = (3x + 2)(2x + 1)
So p(x) = (x − 3)(3x + 2)(2x + 1).
(c) p(eʸ + e⁻ʸ) = 0 means eʸ + e⁻ʸ must equal one of the roots of p: 3, −2/3, or −1/2.
Since eʸ + e⁻ʸ ≥ 2 for all real y (by AM–GM, as eʸ and e⁻ʸ are both positive), it cannot equal the negative values −2/3 or −1/2. So:
eʸ + e⁻ʸ = 3
Let u = eʸ: u + 1/u = 3
u² − 3u + 1 = 0
u = [3 ± √(9−4)]/2 = (3 ± √5)/2
Both values of u are positive, so both are valid for eʸ. …
The full model answer is in the app.
| Session | Paper 2 | Paper 3 |
|---|---|---|
| May/June 2026 | – | 2 |
| May/June 2025 | 2 | 2 |
| October/November 2025 | – | 1 |
| May/June 2024 | 1 | 1 |
| October/November 2024 | 1 | 2 |
| May/June 2023 | 1 | 2 |
| October/November 2023 | 2 | 1 |
| May/June 2022 | – | 2 |
| October/November 2022 | 1 | 2 |
| May/June 2021 | 2 | 1 |
| October/November 2021 | 2 | 1 |
29 real A Level Mathematics (9709) questions on Algebra from 2021 to 2026, from Paper 2, Paper 3. Each one shows the year, session and paper it came from.
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