AQA A-level Mathematics: Polynomial Manipulation, Algebraic Division and the Factor Theorem (7357) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow content statement B6 of the AQA A-level Mathematics 7357 specification (version 1.3, 31 January 2018): expanding and collecting, factorising, algebraic division by linear expressions, the factor theorem and remainders, and rational expressions. Answers show every step, with a bracketed [1] beside each step that would earn credit. Read the study guide first if a method is new, and keep the revision notes beside you. Related pages: course hub, checklist and the free 10-minute diagnostics.

Questions

1. Expand and simplify (x + 4)(x − 2)(x + 5). [3]

2. Factorise completely:

(a) 15x² + 7x − 2 [2] (b) 8x³ − 18x [3]

3. Divide x³ − 5x² + 2x + 9 by (x − 3), giving the quotient and the remainder. [3]

4. When p(x) = 3x³ − kx² + 4x − 5 is divided by (x − 2), the remainder is 7. Find the value of k. [3]

5. The polynomial f(x) is given by f(x) = x³ − 3x² − 10x + 24.

(a) Use the factor theorem to prove that (x − 2) is a factor of f. [2] (b) Hence solve f(x) = 0. [3]

6. The polynomial g(x) = 4x³ + ax² + bx + 3 has (2x + 1) as a factor. The remainder when g(x) is divided by (x − 1) is 9.

(a) Find the values of a and b. [4] (b) Factorise g(x) as far as possible using real numbers, justifying why you stop. [2]

7. (a) Simplify (3x² − 10x − 8)/(x² − 16), stating the values of x that are excluded. [3] (b) Write (x² + 2x − 1)/(x + 3) in the form Ax + B + C/(x + 3). [3]

8. Simplify (x² − 9)/(x² + 2x) × (x + 2)/(x + 3). [3]

9. The polynomial f(x) = x³ + ax² + bx + c has factors (x − 1) and (x + 2). When f(x) is divided by (x − 3) the remainder is −10.

(a) Find the values of a, b and c. [5] (b) Hence factorise f(x) completely and state its roots. [3]

10. A cuboid has volume (2x³ + 9x² + x − 12) cm³, length (x + 4) cm and width (x − 1) cm, where x is a positive constant.

(a) Find an expression for the height of the cuboid. [4] (b) State the smallest range of values that x can take for the cuboid to exist. [1]

11. The polynomial f(x) is given by f(x) = x⁴ + x³ − 7x² − x + 6.

(a) Prove, using the factor theorem, that f has (x + 3) as a factor. [2] (b) Divide f(x) by (x + 3) to find a cubic quotient. [3] (c) Hence factorise f(x) completely. [3]

Answers

1. (x + 4)(x − 2) = x² + 2x − 8 [1] Multiply by (x + 5): x³ + 5x² + 2x² + 10x − 8x − 40 [1] x³ + 7x² + 2x − 40 [1] Examiner insight: Expanding two brackets first and writing the intermediate quadratic makes it easy to see where a slip occurred, and an unsimplified final line leaves marks uncollected.

2. (a) ac = −30, and the numbers 10 and −3 give product −30 and sum 7, so 15x² + 10x − 3x − 2 [1] = 5x(3x + 2) − (3x + 2) = (5x − 1)(3x + 2) [1] (b) Take out the common factor: 2x(4x² − 9) [1] Recognise the difference of two squares: 4x² − 9 = (2x − 3)(2x + 3) [1] 2x(2x − 3)(2x + 3) [1] Examiner insight: “Factorise completely” invites you to take out the common factor first, and an answer that stops at 2x(4x² − 9) has not factorised completely.

3. x² × (x − 3) = x³ − 3x², leaving −2x² + 2x [1] −2x × (x − 3) = −2x² + 6x, leaving −4x + 9; then −4 × (x − 3) = −4x + 12, leaving −3 [1] Quotient x² − 2x − 4, remainder −3 [1] Examiner insight: State clearly which expression is the quotient and which is the remainder, since a reader has to be able to find both. Checking f(3) = 27 − 45 + 6 + 9 = −3 confirms the remainder.

4. Remainder theorem: p(2) = 7 [1] 24 − 4k + 8 − 5 = 7, so 27 − 4k = 7 [1] k = 5 [1] Examiner insight: Name the theorem or write “p(2) = 7” explicitly. A bare value with no substitution shown is hard to give credit to if the arithmetic goes wrong.

5. (a) f(2) = 8 − 12 − 20 + 24 [1] = 0, so (x − 2) is a factor by the factor theorem [1] (b) Divide: f(x) = (x − 2)(x² − x − 12) [1] x² − x − 12 = (x − 4)(x + 3) [1] x = 2, x = 4, x = −3 [1] Examiner insight: In a “show that” part, show the substitution line and state the conclusion in words. For “hence”, use the factor you were given rather than starting again.

6. (a) g(−1/2) = 0: −1/2 + a/4 − b/2 + 3 = 0, which gives a − 2b = −10 [1] g(1) = 9: 4 + a + b + 3 = 9, which gives a + b = 2 [1] Subtract: 3b = 12, so b = 4 [1] a = −2, b = 4 [1] (b) g(x) = (2x + 1)(2x² − 2x + 3) [1] Discriminant of the quadratic: (−2)² − 4(2)(3) = −20 < 0, so it has no real factors [1] Examiner insight: Substituting x = −1/2 for the factor (2x + 1) is the step most often done wrongly. Keep fractions exact and check your answer by testing both conditions in the original polynomial.

7. (a) 3x² − 10x − 8 = (3x + 2)(x − 4) [1] x² − 16 = (x − 4)(x + 4) [1] (3x + 2)/(x + 4), with x ≠ 4 and x ≠ −4 [1] (b) (x + 3)(x − 1) = x² + 2x − 3, so x² + 2x − 1 is 2 more [1] Quotient x − 1 and remainder 2 [1] x − 1 + 2/(x + 3) [1] Examiner insight: Cancel only after both parts are in factorised form. Where a question asks for excluded values, take them from the original denominator, not from the simplified one.

8. x² − 9 = (x − 3)(x + 3) and x² + 2x = x(x + 2) [1] (x − 3)(x + 3)/[x(x + 2)] × (x + 2)/(x + 3) [1] Cancelling (x + 3) and (x + 2) gives (x − 3)/x [1] Examiner insight: Factorising every numerator and denominator before multiplying keeps the cancelling visible. Cancelling across the multiplication sign without factorising invites errors.

9. (a) f(1) = 0: 1 + a + b + c = 0 [1] f(−2) = 0: −8 + 4a − 2b + c = 0 [1] f(3) = −10: 27 + 9a + 3b + c = −10 [1] Subtracting the first from the second and third: a − b = 3 and 4a + b = −18 [1] Adding gives 5a = −15, so a = −3, b = −6, c = 8 [1] (b) The third factor is (x − k), and the constant term gives (−1)(2)(−k) = 8, so k = 4 [1] f(x) = (x − 1)(x + 2)(x − 4) [1] Roots: x = 1, x = −2, x = 4 [1] Examiner insight: Number your three equations and say which fact produced each. When elimination goes wrong, a clear list of equations lets a reader see which method steps were sound.

10. (a) Volume = length × width × height, so height = V ÷ (x + 4) ÷ (x − 1) [1] Dividing 2x³ + 9x² + x − 12 by (x + 4) gives 2x² + x − 3 with remainder 0 [1] 2x² + x − 3 = (2x + 3)(x − 1) [1] Height = (2x + 3) cm [1] (b) The width must be positive, so x > 1 [1] Examiner insight: In a context question, finish with the answer in the units of the problem. A restriction on x comes from the physical meaning, which here is a positive width.

11. (a) f(−3) = 81 − 27 − 63 + 3 + 6 [1] = 0, so (x + 3) is a factor [1] (b) x³ × (x + 3) = x⁴ + 3x³, leaving −2x³ − 7x²; then −2x² × (x + 3) = −2x³ − 6x², leaving −x² − x [1] −x × (x + 3) = −x² − 3x, leaving 2x + 6; then 2 × (x + 3) = 2x + 6, leaving 0 [1] Quotient x³ − 2x² − x + 2 [1] (c) x³ − 2x² − x + 2 = x²(x − 2) − (x − 2) = (x − 2)(x² − 1) [1] = (x − 2)(x − 1)(x + 1) [1] f(x) = (x + 3)(x − 2)(x − 1)(x + 1) [1] Examiner insight: Include the zero-coefficient columns in a quartic division, and do not forget to carry the earlier factor into the final answer when “hence factorise f(x)” is the instruction.

Where marks are usually lost

• Using f(a) for a factor (x + a) instead of f(−a).
• For a divisor (2x + 1), substituting x = −1 instead of x = −1/2.
• Dropping a negative sign when subtracting during long division.
• Omitting a placeholder zero when a power is missing from the dividend.
• Stopping at one root or one factor when the question asks for all roots or “completely”.
• Cancelling terms rather than factors in a fraction.
• Leaving out excluded values when asked for them.
• Presenting a remainder that still contains x after dividing by a linear expression.

Next steps

• Review the methods in the revision notes.
• Re-read the study guide for any method you missed.
• Visit the AQA A-level Mathematics hub and tick progress on the checklist.
• Try the free 10-minute diagnostics to find your weakest area.
• Tutor help: Book a free trial class if you would like it.

Official syllabus

AQA A-level Mathematics (7357) specification, for teaching from September 2017, version 1.3 dated 31 January 2018, published by AQA. These questions cover content statement B6.