A-Level Mathematics October/November 2024 Q1: The polynomial 4x³ + ax² +5x+b, where a and b are constants, is denoted by p(x). It is…

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  Let $p(x) = 4x^3 + ax^2 + 5x + b$.

  It is given that $(2x+1)$ is a factor of $p(x)$. By the Factor Theorem, if $(2x+1)$ is a factor, then $p(-\frac{1}{2}) = 0$.

  $p(-\frac{1}{2}) = 4(-\frac{1}{2})^3 + a(-\frac{1}{2})^2 + 5(-\frac{1}{2}) + b = 0$ $4(-\frac{1}{8}) + a(\frac{1}{4}) - \frac{5}{2} + b = 0$ $-\frac{1}{2} + \frac{a}{4} - \frac{5}{2} + b = 0$ $\frac{a}{4} + b - 3 = 0$ $a + 4b = 12 \quad \cdots (1)$

  It is also given that when $p(x)$ is divided by $(x-4)$, the remainder is 3 times the remainder when $p(x)$ is divided by $(x-2)$. By the Remainder Theorem, the remainder when $p(x)$ is divided by $(x-c)$ is $p(c)$. So, $p(4) = 3 \times p(2)$.

  Remainder $p(4) = 4(4)^3 + a(4)^2 + 5(4) + b$ $p(4) = 4(64) + 16a + 20 + b$ $p(4) = 256 + 16a + 20 + b = 276 + 16a + b$

  Remainder $p(2) = 4(2)^3 + a(2)^2 + 5(2) + b$ $p(2) = 4(8) + 4a + 10 + b$ $p(2) = 32 + 4a + 10 + b = 42 + 4a + b$

  Now, form the equation $p(4) = 3p(2)$: $276 + 16a + b = 3(42 + 4a + b)$ $276 + 16a + b = 126 + 12a + 3b$ $276 - 126 = 12a - 16a + 3b - b$ $150 = -4a + 2b$ Dividing by 2 gives: $75 = -2a + b \quad \cdots (2)$

  Now we solve the simultaneous equations (1) and (2). From (2), $b = 75 + 2a$. Substitute this into (1): $a + 4(75 + 2a) = 12$ $a + 300 + 8a = 12$ $9a = 12 - 300$ $9a = -288$ $a = -32$

  Substitute $a = -32$ into $b = 75 + 2a$: $b = 75 + 2(-32)$ $b = 75 - 64$ $b = 11$

  Therefore, the values are $a = -32$ and $b = 11$.

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  How the marks are awarded

  • M1 — Applying the Factor Theorem by substituting x = -1/2 into p(x) and equating the expression to zero.
  • A1 — Obtaining a correct linear equation in a and b from the Factor Theorem, such as 'a + 4b = 12'.
  • M1 — Applying the Remainder Theorem by finding expressions for p(4) and p(2) and forming the equation p(4) = 3p(2).
  • A1 — Obtaining a second correct linear equation in a and b from the remainder condition, such as '75 = -2a + b'.
  • A1 — Correctly solving the pair of simultaneous linear equations to find the final, correct values for both a and b.

  Common mistakes

  • Making a sign error when substituting x = -1/2, for example calculating (-1/2)³ as +1/8 or (-1/2)² as -1/4.
  • Incorrectly applying the Factor Theorem for the factor (2x+1), for instance by substituting x = -1 instead of x = -1/2.
  • Making an algebraic error when expanding the bracket in the remainder condition, e.g., writing 3(42 + 4a + b) as 126 + 12a + b, forgetting to multiply the 'b' term by 3.
  • Errors in the process of solving the simultaneous equations, such as substitution mistakes or arithmetic errors during elimination.

  Examiner tip: Master the Factor and Remainder Theorems to translate statements about polynomials and their divisors into algebraic equations that can be solved simultaneously.

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