OCR C2 2010 January — Question 6 9 marks

Exam BoardOCR
ModuleC2 (Core Mathematics 2)
Year2010
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFactor & Remainder Theorem
TypeSingle polynomial, two remainder/factor conditions
DifficultyModerate -0.3 This is a standard C2 factor/remainder theorem question requiring students to set up two simultaneous equations (f(-3)=0 and f(2)=35) and solve for a and b, then perform polynomial division. While it involves multiple steps, the techniques are routine and commonly practiced at this level, making it slightly easier than average.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02k Simplify rational expressions: factorising, cancelling, algebraic division

6 The cubic polynomial \(\mathrm { f } ( x )\) is given by $$\mathrm { f } ( x ) = 2 x ^ { 3 } + a x ^ { 2 } + b x + 15$$ where \(a\) and \(b\) are constants. It is given that ( \(x + 3\) ) is a factor of \(\mathrm { f } ( x )\) and that, when \(\mathrm { f } ( x )\) is divided by ( \(x - 2\) ), the remainder is 35 .
  1. Find the values of \(a\) and \(b\).
  2. Using these values of \(a\) and \(b\), divide \(\mathrm { f } ( x )\) by ( \(x + 3\) ).

AnswerMarks Guidance
(i)\(f(-3) = 0 \Rightarrow -54 + 9a - 3b + 15 = 0\) M1
\(3a - b = 13\)A1 Obtain \(3a - b = 13\), or unsimplified equiv
\(f(2) = 35 \Rightarrow 16 + 4a + 2b + 15 = 35\)M1 Attempt f(2) and equate to 35, or equiv method
\(2a + b = 2\)A1 Obtain \(2a + b = 2\), or unsimplified equiv
Hence \(a = 3, b = -4\)M1, A1 Attempt to solve simultaneous eqns; Obtain \(a = 3, b = -4\)
(ii)\(f(x) = (x+3)(2x^2 - 3x + 5)\) M1, A1
ie quotient is \((2x^2 - 3x + 5)\)A1 Obtain \(2x^2 - 3x + 5\) (state or imply as quotient)
(i) | $f(-3) = 0 \Rightarrow -54 + 9a - 3b + 15 = 0$ | M1 | Attempt f(-3) and equate to 0, or equiv method |
| $3a - b = 13$ | A1 | Obtain $3a - b = 13$, or unsimplified equiv |
| $f(2) = 35 \Rightarrow 16 + 4a + 2b + 15 = 35$ | M1 | Attempt f(2) and equate to 35, or equiv method |
| $2a + b = 2$ | A1 | Obtain $2a + b = 2$, or unsimplified equiv |
| Hence $a = 3, b = -4$ | M1, A1 | Attempt to solve simultaneous eqns; Obtain $a = 3, b = -4$ |

(ii) | $f(x) = (x+3)(2x^2 - 3x + 5)$ | M1, A1 | Attempt complete division by $(x+3)$, or equiv; Obtain $2x^2 - 3x + c$ or $2x^2 + bx + 5$, from correct f(x) |
| ie quotient is $(2x^2 - 3x + 5)$ | A1 | Obtain $2x^2 - 3x + 5$ (state or imply as quotient) |
6 The cubic polynomial $\mathrm { f } ( x )$ is given by

$$\mathrm { f } ( x ) = 2 x ^ { 3 } + a x ^ { 2 } + b x + 15$$

where $a$ and $b$ are constants. It is given that ( $x + 3$ ) is a factor of $\mathrm { f } ( x )$ and that, when $\mathrm { f } ( x )$ is divided by ( $x - 2$ ), the remainder is 35 .\\
(i) Find the values of $a$ and $b$.\\
(ii) Using these values of $a$ and $b$, divide $\mathrm { f } ( x )$ by ( $x + 3$ ).

\hfill \mbox{\textit{OCR C2 2010 Q6 [9]}}