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A-LevelMathematicsAlgebra9709/21 · May/June 2025 · Q59 Marks

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).

  1. (a)Find the values of a and b.[4]
  2. (b)Hence factorise p(x) completely.[3]
  3. (c)Hence solve the equation p(3cosecθ) = 0 for 90° < θ < 270°.[2]

Mark scheme

Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.

  1. [M1] Substitute x = 3/2 equating to 0, and x = −1 equating to -15.
  2. [A1] Obtain 27/8 a + 9/4 b - 3/2 a - 24 = 0 (OE, Powers of 3/2 must be evaluated).
  3. [A1] Obtain -a+b+a-24=-15.
  4. [A1] Solve to obtain a = 2 and b = 9.
  5. [M1] Divide p(x) by 2x-3 (OE method, such as inspection. For algebraic long division, must go as far as the term in x. Allow attempt at synthetic division.).
  6. [A1] Obtain quotient x² +6x+8 (Allow 2x² +12x+16 from synthetic division).
  7. [A1] Conclude (2x-3)(x+2)(x+4).
  8. [M1] Use factorised form to determine value of sinθ, with -1<sinθ<1.
  9. [A1] Obtain finally sin θ = -1/2 only and hence angle 228.6 (Or greater accuracy 228.590... Ignore solutions outside the range).

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About this question

Topic
Algebra, A-Level Mathematics (syllabus 9709). Structured question, 9 marks.
Source
A-Level Mathematics, May/June 2025 Paper 2 Variant 1, question 5.

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