OxfordAQA International AS and A-level Mathematics: Polynomials, factor theorem and graphs (9660) – 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 OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, content area P1.1: polynomials, the Remainder Theorem and Factor Theorem, sketching, intersections and transformations. This content is part of International AS, Unit P1 (Pure Maths), and the specification gives no Core, Extension or Higher labels. It applies to International AS exams from May/June 2018 and International A-level exams from May/June 2019. Marks in square brackets are Marlbridge’s own guide to the size of each part.

Links: study guide, revision notes, OxfordAQA Mathematics hub, printable checklist, Unit P1 practice and the free 10-minute diagnostics.

Questions

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

2. Find the remainder when 4x³ − 3x + 7 is divided by (x + 2). [2]

3. f(x) = x³ − 3x² − 6x + 8.

(a) Show that (x − 1) is a factor of f(x). [1] (b) Factorise f(x) completely. [3] (c) Solve f(x) = 0. [1]

4. The polynomial g(x) = 2x³ + px² + qx − 6 has (x − 1) as a factor. When g(x) is divided by (x + 1), the remainder is 12. Find the values of p and q. [5]

5. h(x) = x³ − 3x² + 4x − 2. Show that the equation h(x) = 0 has exactly one real root, and state its value. [5]

6. Sketch the curve y = (x − 4)²(x + 1). Label every point where it meets the axes. [4]

7. Sketch the curve y = −x² + 4x + 5. Label the axis intercepts, the line of symmetry and the vertex. [4]

8. The curve C has equation y = (x − 4)²(x + 1) and the line L has equation y = x + 16.

(a) Show that the x-coordinates of the points where C meets L satisfy x³ − 7x² + 7x = 0. [2] (b) Find the exact x-coordinates of the points of intersection. [3] (c) Find the exact y-coordinate of the intersection point with the largest x-coordinate. [1]

9. Describe fully the single transformation that maps the graph of y = f(x) onto each of the following.

(a) y = f(x) − 5 [1] (b) y = f(x − 2) [1] (c) y = 3f(x) [1] (d) y = f(x/2) [1] (e) y = f(−x) [1]

10. The curve P has equation y = f(x), where f(x) = x² − 3x + 1. Find the equation of the image of P after each single transformation.

(a) Translation by the vector (2, −3). [2] (b) Stretch parallel to the y-axis, scale factor 3. [1] (c) Stretch parallel to the x-axis, scale factor 1/2. [1] (d) Reflection in the x-axis. [1]

11. f(x) = (x + 2)(x − 1)(x − 3).

(a) Sketch y = f(x), labelling all intercepts. [3] (b) Describe the transformation that maps y = f(x) onto y = f(x + 1), and write down the x-coordinates where the new curve meets the x-axis. [2] (c) Write down the y-intercept of the curve y = 2f(x). [1]

12. f(x) = x³ + x² − 16x + 20.

(a) Show that (x − 2) is a factor of f(x), and factorise f(x) completely. [4] (b) Sketch y = f(x), labelling the intercepts. [3] (c) The line y = 20 meets the curve y = f(x) at three points. Find the exact x-coordinates. [3] (d) The curve is translated by the vector (3, 0). Write down the x-coordinates where the new curve meets the x-axis. [2]

Answers

1. (x − 5)(x + 1) = x² − 4x − 5 [1] (3x + 2)(x² − 4x − 5) = 3x³ − 12x² − 15x + 2x² − 8x − 10 [1] = 3x³ − 10x² − 23x − 10 [1] [3] Examiner insight: Expanding two brackets first and then multiplying by the third keeps the number of terms manageable, and a final check that there are four terms catches many slips.

2. Use a = −2: f(−2) = 4(−8) − 3(−2) + 7 [1] = −32 + 6 + 7 = −19 [1] [2] Examiner insight: A divisor of (x + 2) needs the value −2, not 2. Showing the substitution lets a sign slip in the arithmetic be separated from a wrong value of a.

3. (a) f(1) = 1 − 3 − 6 + 8 = 0, so (x − 1) is a factor [1] (b) Divide: x³ − 3x² − 6x + 8 = (x − 1)(x² − 2x − 8) [1] (check the x term: −8 + 2 = −6, as required) x² − 2x − 8 = (x − 4)(x + 2) [1] f(x) = (x − 1)(x − 4)(x + 2) [1] (c) x = 1, 4 or −2 [1] [5] Examiner insight: A “show that” part needs the substitution written out and the conclusion stated, not just the number 0. In part (c), give all three roots, taking the sign change from each factor.

4. g(1) = 0: 2 + p + q − 6 = 0, so p + q = 4 [1] g(−1) = 12: −2 + p − q − 6 = 12, so p − q = 20 [1] Add the equations: 2p = 24 [1] p = 12 [1] q = −8 [1] [5] Examiner insight: Write each condition as an equation before solving. Checking g(1) = 2 + 12 − 8 − 6 = 0 at the end is a quick way to catch a sign error.

5. h(1) = 1 − 3 + 4 − 2 = 0, so (x − 1) is a factor [1] h(x) = (x − 1)(x² − 2x + 2) [1] For x² − 2x + 2, discriminant = (−2)² − 4(1)(2) [1] = −4, which is negative, so no real roots [1] So the only real root is x = 1 [1] [5] Examiner insight: “Exactly one” needs both parts: a root found, and a reason the quadratic factor gives no more. Quoting the discriminant value alongside the word negative makes the reasoning easy to credit.

6. y-intercept: (0, 16) [1] x-intercepts: (−1, 0) where the curve crosses [1] (4, 0) where the curve touches the axis (repeated factor) [1] Positive cubic shape: from lower left, up to a peak between −1 and 4, down to touch at (4, 0), then up to the upper right [1] [4] Examiner insight: A repeated root drawn as a crossing is a common loss of credit. The curve should visibly turn on the x-axis at x = 4, and the intercepts should be labelled as coordinates.

7. Roots: −(x − 5)(x + 1) = 0, so x = 5 and x = −1 [1] y-intercept (0, 5) [1] Line of symmetry x = 2, vertex (2, 9) [1] ∩-shaped parabola through these points, with the vertex labelled [1] [4] Examiner insight: The sign of the x² coefficient decides the shape. The line of symmetry is the midpoint of the roots, which also gives the x-coordinate of the vertex.

8. (a) (x − 4)²(x + 1) = (x² − 8x + 16)(x + 1) = x³ − 7x² + 8x + 16 [1] Set equal to x + 16: x³ − 7x² + 7x = 0 [1] (b) x(x² − 7x + 7) = 0, so x = 0 [1] x = (7 ± √(49 − 28))/2 [1] x = 0, (7 − √21)/2, (7 + √21)/2 [1] (c) y = x + 16 with x = (7 + √21)/2 gives y = (39 + √21)/2 [1] [6] Examiner insight: Dividing the equation through by x loses a solution. Take the common factor out instead, and give surd answers exactly if the question says “exact”.

9. (a) Translation by the vector (0, −5) [1] (b) Translation by the vector (2, 0) [1] (c) Stretch parallel to the y-axis, scale factor 3 [1] (d) Stretch parallel to the x-axis, scale factor 2 [1] (e) Reflection in the y-axis [1] [5] Examiner insight: A full description names the type (translation, stretch or reflection) and the detail: a vector, or a direction with a scale factor, or the line of reflection. Writing only “stretch” or “shift” usually leaves out part of what is asked.

10. (a) Replace x by (x − 2) and subtract 3: (x − 2)² − 3(x − 2) + 1 − 3 [1] y = x² − 7x + 8 [1] (b) y = 3x² − 9x + 3 [1] (c) Replace x by 2x: y = 4x² − 6x + 1 [1] (d) y = −x² + 3x − 1 [1] [5] Examiner insight: Multiply every term for a vertical stretch or a reflection in the x-axis, and substitute into every x for horizontal changes. Applying a stretch to only some terms is a typical slip.

11. (a) Roots at x = −2, 1, 3, all crossing [1] y-intercept: f(0) = (2)(−1)(−3) = 6, so (0, 6) [1] Positive cubic shape through these points [1] (b) Translation by the vector (−1, 0) [1] Roots at x = −3, 0, 2 [1] (c) (0, 12) [1] [6] Examiner insight: For y = f(x + 1), each root moves one unit to the left, so subtract 1 from each original root. Checking that x = 0 now gives a root, since the old root x = 1 has moved there, confirms the answer.

12. (a) f(2) = 8 + 4 − 32 + 20 = 0, so (x − 2) is a factor [1] f(x) = (x − 2)(x² + 3x − 10) [1] x² + 3x − 10 = (x − 2)(x + 5) [1] f(x) = (x − 2)²(x + 5) [1] (b) Intercepts: (2, 0) touching, (−5, 0) crossing [1] y-intercept (0, 20) [1] Positive cubic: from lower left, crossing at −5, peak, touching the axis at 2, then rising [1] (c) x³ + x² − 16x + 20 = 20 gives x³ + x² − 16x = 0 [1] x(x² + x − 16) = 0, so x = 0 [1] x = (−1 ± √65)/2, so x = 0, (−1 − √65)/2, (−1 + √65)/2 [1] (d) Replace x by (x − 3): roots at x − 3 = 2 and x − 3 = −5 [1] x = 5 and x = −2 [1] [12] Examiner insight: On a multi-part problem the later parts often reuse the factorisation. A correct factorised form lets you read off roots for part (d) without expanding the translated cubic.

Where marks are usually lost

• Testing the wrong value of a: x + 2 as a divisor needs f(−2).
• Writing “(x − 2) is a factor” without showing f(2) = 0.
• Stopping at a quadratic factor without checking whether it factorises or has a negative discriminant.
• Dividing by x in an intersection equation and losing the root x = 0.
• Showing a repeated root as a crossing, or missing out the y-intercept.
• Moving f(x − 2) to the left, or f(2x) to scale factor 2.
• Describing a transformation without its direction, vector or scale factor.
• Transforming only some terms of the equation when asked for the image curve.
• Leaving intercepts as single numbers when coordinates are asked for.

Next steps

• Revise with the revision notes.
• Read the study guide for any question you missed.
• Go to the OxfordAQA Mathematics hub and the printable checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA. See section 3.1, International AS Unit P1 (Pure Maths), content area P1.1.