OCR A Level Mathematics A: Polynomials, factor theorem, rational expressions, partial fractions and modelling with functions (H240) – 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 the OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), content references 1.02j, 1.02k, 1.02y and 1.02z. Questions 1, 2, 3 and 5 use Stage 1 skills (common with AS). The remaining questions use A Level only content: division with a quotient and remainder, quartics, rational expressions, partial fractions and modelling. Read the study guide first and use the revision notes for recall. See also the course hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. Expand and simplify (x − 4)(x² + 3x − 2) − x(x + 1). [2]

2. The polynomial f(x) = x³ + ax − 10 has a factor (x − 2). Find a, and show that f(x) = 0 has only one real root. [4]

3. The polynomial p(x) = x³ + ax² + bx − 6 has a factor (x − 1). The remainder when p(x) is divided by (x + 2) is −12. Find a and b. [4]

4. Divide 4x³ + 4x² − 15x + 7 by (2x + 3), stating the quotient and the remainder. [4]

5. g(x) = 3x³ − 4x² − 13x − 6.

(a) Show that (3x + 2) is a factor of g(x). [2] (b) Factorise g(x) completely. [3] (c) Write down the solutions of g(x) = 0. [1]

6. h(x) = x⁴ − x³ − 19x² + 49x − 30.

(a) Show that (x − 2) and (x + 5) are both factors of h(x). [2] (b) Hence solve h(x) = 0. [4]

7. Simplify (2x² + 5x − 12)/(x² + x − 12), and state the values of x for which the original expression is undefined. [4]

8. Express (6x + 11)/[(2x − 1)(x + 3)] in partial fractions. [5]

9. Express (11 − 7x)/[(x + 3)(x − 1)²] in the form A/(x + 3) + B/(x − 1) + C/(x − 1)². [6]

10.

(a) Express (4x² + 8x − 3)/[(x − 1)(x + 2)] in the form P + A/(x − 1) + B/(x + 2), where P, A and B are constants. [4] (b) Hence find the exact value of the integral of (4x² + 8x − 3)/[(x − 1)(x + 2)] from x = 2 to x = 3, giving your answer in the form a + ln b. [3]

11. An open tank has a square base of side x metres and a volume of 4 m³.

(a) Show that the outside surface area is S = x² + 16/x square metres. [3] (b) The tank is to have a surface area of 20 m². Show that x³ − 20x + 16 = 0. [2] (c) Solve the equation in (b), giving exact values, and state which values are acceptable. [4] (d) State one assumption made in the model for S. [1]

12. The heights of a seedling, in centimetres, were recorded at four times t, in weeks.

t	0	2	4	6
Height	10	14	15	13

Model L is h = 10 + 1.5t. Model Q is h = 10 + 3t − 0.5t².

(a) Complete the model values at each time for both models. [2] (b) Compare the two models using the total of the differences from the recorded heights, ignoring sign. [2] (c) Use Model Q to find when the height would return to zero, correct to 3 significant figures. [2] (d) State one reason why Model Q is unsuitable for large t. [1]

Answers

1. (x − 4)(x² + 3x − 2) = x³ + 3x² − 2x − 4x² − 12x + 8 = x³ − x² − 14x + 8 [1] Subtract x² + x, with both signs changed: x³ − 2x² − 15x + 8 [1] [2] Examiner insight: the sign change on the second bracket is the usual slip, so write the expanded second product in brackets before subtracting.

2. f(2) = 8 + 2a − 10 = 0 [1] So a = 1 [1] f(x) = x³ + x − 10 = (x − 2)(x² + 2x + 5) [1] For x² + 2x + 5 the discriminant is 4 − 20 = −16 < 0, so there are no real roots and the only real root is x = 2 [1] [4] Examiner insight: the second part asks you to justify “only one”, so a statement about the discriminant of the quadratic factor is what answers it.

3. p(1) = 1 + a + b − 6 = 0, so a + b = 5 [1] p(−2) = −8 + 4a − 2b − 6 = −12, so 2a − b = 1 [1] Add the equations: 3a = 6 [1] a = 2, b = 3 [1] [4] Examiner insight: write each substitution as an equation before simplifying, which keeps the working easy to check and makes slips easier to find.

4. 4x³ ÷ 2x = 2x², subtract 4x³ + 6x² to leave −2x² − 15x [1] −2x² ÷ 2x = −x, subtract −2x² − 3x to leave −12x + 7 [1] −12x ÷ 2x = −6, subtract −12x − 18 to leave 25 [1] Quotient 2x² − x − 6, remainder 25 [1] [4] Examiner insight: check the remainder by evaluating the polynomial at x = −3/2, which gives 25. A matching check is quick and is good practice on any division.

5. (a) g(−2/3) = 3(−8/27) − 4(4/9) − 13(−2/3) − 6 [1] = −8/9 − 16/9 + 78/9 − 54/9 = 0, so (3x + 2) is a factor [1] (b) g(x) = (3x + 2)(x² − 2x − 3) [1] x² − 2x − 3 = (x − 3)(x + 1) [1] g(x) = (3x + 2)(x − 3)(x + 1) [1] (c) x = −2/3, 3, −1 [1] [6] Examiner insight: for a bracket (3x + 2) the value to substitute is −2/3, and a conclusion sentence, “so (3x + 2) is a factor”, is part of a “show that”.

6. (a) h(2) = 16 − 8 − 76 + 98 − 30 = 0 [1] h(−5) = 625 + 125 − 475 − 245 − 30 = 0, so both are factors [1] (b) (x − 2)(x + 5) = x² + 3x − 10, so divide h(x) by x² + 3x − 10 [1] The quotient is x² − 4x + 3, because (x² + 3x − 10)(x² − 4x + 3) = x⁴ − x³ − 19x² + 49x − 30 [1] x² − 4x + 3 = (x − 1)(x − 3) [1] x = −5, 1, 2, 3 [1] [6] Examiner insight: “hence” means use the factors you have just shown, so dividing by the product of the two brackets is the efficient route and avoids extra searching for roots.

7. 2x² + 5x − 12 = (2x − 3)(x + 4) [1] x² + x − 12 = (x + 4)(x − 3) [1] Cancel (x + 4): (2x − 3)/(x − 3) [1] The original is undefined where x² + x − 12 = 0, so x = 3 and x = −4 [1] [4] Examiner insight: cancel brackets, not terms, and quote the exclusions from the original denominator, because x = −4 is excluded even though the simplified form allows it.

8. 6x + 11 ≡ A(x + 3) + B(2x − 1) [1] x = 1/2: 14 = (7/2)A [1] A = 4 [1] x = −3: −7 = −7B [1] B = 1, so the result is 4/(2x − 1) + 1/(x + 3) [1] [5] Examiner insight: substituting the value that makes one bracket zero is faster than comparing coefficients. A check at a value you have not used, such as x = 1, gives 17/4 on both sides and takes a few seconds.

9. 11 − 7x ≡ A(x − 1)² + B(x + 3)(x − 1) + C(x + 3) [1] x = −3: 32 = 16A, so A = 2 [1] x = 1: 4 = 4C, so C = 1 [1] Compare x² terms: 0 = A + B [1] B = −2 [1] Result: 2/(x + 3) − 2/(x − 1) + 1/(x − 1)² [1] [6] Examiner insight: the repeated factor needs both B/(x − 1) and C/(x − 1)². Leaving out the middle term makes the identity impossible to satisfy.

10. (a) The denominator is x² + x − 2 [1] 4(x² + x − 2) = 4x² + 4x − 8, so the remainder is 4x + 5 and P = 4 [1] 4x + 5 ≡ A(x + 2) + B(x − 1): x = 1 gives 9 = 3A, so A = 3 [1] x = −2 gives −3 = −3B, so B = 1: 4 + 3/(x − 1) + 1/(x + 2) [1] (b) Antiderivative: 4x + 3 ln|x − 1| + ln|x + 2| [1] Limits: (12 + 3 ln 2 + ln 5) − (8 + 0 + ln 4) [1] = 4 + 3 ln 2 + ln 5 − 2 ln 2 = 4 + ln 10 [1] [7] Examiner insight: both numerator and denominator have degree 2, so divide first. Combine the logarithms at the end, because the question asks for the form a + ln b.

11. (a) Height h: x²h = 4, so h = 4/x² [1] Surface area = base + four sides = x² + 4xh [1] = x² + 4x(4/x²) = x² + 16/x [1] (b) x² + 16/x = 20; multiply by x: x³ + 16 = 20x [1] So x³ − 20x + 16 = 0 [1] (c) At x = 4: 64 − 80 + 16 = 0, so (x − 4) is a factor [1] (x − 4)(x² + 4x − 4) = 0 [1] x = (−4 ± √32)/2 = −2 ± 2√2 [1] Length cannot be negative, so x = 4 or x = 2√2 − 2 (about 0.828) [1] (d) The walls have negligible thickness, or the joins use no extra material [1] [10] Examiner insight: in a “show that”, write the volume relation and the substitution that eliminates h, then reach the stated result. Reject the negative root explicitly, with the reason from the context.

12. (a) Model L: 10, 13, 16, 19 [1] Model Q: 10, 14, 14, 10 [1] (b) Differences for L: 0, 1, 1, 6 with total 8 [1] Differences for Q: 0, 0, 1, 3 with total 4, so Model Q fits better over this range [1] (c) 10 + 3t − 0.5t² = 0, so t² − 6t − 20 = 0 and t = 3 ± √29 [1] Positive root: t = 8.39 weeks [1] (d) A seedling’s height does not fall back to zero, so the model fails for large t [1] [7] Examiner insight: “compare” needs a stated criterion and a conclusion that follows from it, so give the totals and say which model is closer.

Where marks are usually lost

• Substituting the number inside the bracket, such as x = 2 for (2x − 1), instead of its root.
• Dropping a sign when subtracting a product or a line in long division.
• Not writing a conclusion in a “show that” question.
• Missing the B/(x − 1) term when the denominator contains a squared factor.
• Using partial fractions without dividing first when the top is of equal or higher degree.
• Leaving an integral as separate logarithms when the question asks for a single form.
• Cancelling terms instead of factors in a rational expression.
• Giving a root of a model’s equation that lies outside the sensible domain.
• Saying a model is “better” with no criterion or figures.

Next steps

• Recall the key results in the revision notes.
• Re-read the explanations in the study guide.
• Return to the course hub and tick off the printable checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR A Level Mathematics A (H240) specification, Version 3.1 (August 2026), for first assessment in 2018, published by OCR. These questions cover content references 1.02j, 1.02k, 1.02y and 1.02z.