Question 5 - A-Level Maths

Edexcel C3 — Question 5 12 marks

Exam Board	Edexcel
Module	C3 (Core Mathematics 3)
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Factor & Remainder Theorem
Type	Verify factor then simplify rational expression
Difficulty	Standard +0.3 This is a structured multi-part question requiring factor theorem verification (routine substitution), polynomial division, rational expression simplification, and quotient rule differentiation. While part (c) involves finding stationary points of a rational function, the algebraic simplification in part (b) makes the calculus straightforward. All techniques are standard C3 content with no novel problem-solving required, making it slightly easier than average.
Spec	1.02j Manipulate polynomials: expanding, factorising, division, factor theorem 1.02k Simplify rational expressions: factorising, cancelling, algebraic division 1.07n Stationary points: find maxima, minima using derivatives 1.07q Product and quotient rules: differentiation

5. (a) Show that $( 2 x + 3 )$ is a factor of $\left( 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 \right)$.
(b) Hence, simplify $$\frac { 2 x ^ { 2 } + x - 3 } { 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 } .$$ (c) Find the coordinates of the stationary points of the curve with equation $$y = \frac { 2 x ^ { 2 } + x - 3 } { 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 } .$$

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Part (a)

Answer	Marks
let $f(x) = 2x^3 - x^2 + 4x + 15$	M1 A1
$f(-\frac{3}{2}) = -\frac{27}{4} - \frac{9}{4} - 6 + 15 = 0$ ∴ $(2x+3)$ is a factor

Part (b)

Answer	Marks	Guidance
\(\frac{x^2 - 2x + 5}{2x + 3} \begin{array}{	c} 2x^3 - x^2 + 4x + 15 \\ 2x^3 + 3x^2 \\ -4x^2 + 4x \\ -4x^2 - 6x \\ 10x + 15 \\ 10x + 15 \end{array}\)	M1 A1
∴ $f(x) = (2x+3)(x^2 - 2x + 5)$
$\frac{2x^2 + x - 3}{2x^3 - x^2 + 4x + 15} = \frac{(2x+3)(x-1)}{(2x+3)(x^2-2x+5)} = \frac{x-1}{x^2-2x+5}$	M1 A1

Part (c)

Answer	Marks
$\frac{dy}{dx} = \frac{1x(x^2-2x+5)-(x-1)(2x-2)}{(x^2-2x+5)^2} = \frac{-x^2+2x+3}{(x^2-2x+5)^2}$	M1 A2
SP: $\frac{-x^2+2x+3}{(x^2-2x+5)^2} = 0$	M1
$-x^2 + 2x + 3 = 0, -(x+1)(x-3) = 0$
$x = -1, 3$	M1 A2
(12)

**Part (a)**
let $f(x) = 2x^3 - x^2 + 4x + 15$ | M1 A1 |
$f(-\frac{3}{2}) = -\frac{27}{4} - \frac{9}{4} - 6 + 15 = 0$ ∴ $(2x+3)$ is a factor |

**Part (b)**
$\frac{x^2 - 2x + 5}{2x + 3} \begin{array}{|c} 2x^3 - x^2 + 4x + 15 \\ 2x^3 + 3x^2 \\ -4x^2 + 4x \\ -4x^2 - 6x \\ 10x + 15 \\ 10x + 15 \end{array}$ | M1 A1 |
∴ $f(x) = (2x+3)(x^2 - 2x + 5)$ |
$\frac{2x^2 + x - 3}{2x^3 - x^2 + 4x + 15} = \frac{(2x+3)(x-1)}{(2x+3)(x^2-2x+5)} = \frac{x-1}{x^2-2x+5}$ | M1 A1 |

**Part (c)**
$\frac{dy}{dx} = \frac{1x(x^2-2x+5)-(x-1)(2x-2)}{(x^2-2x+5)^2} = \frac{-x^2+2x+3}{(x^2-2x+5)^2}$ | M1 A2 |
SP: $\frac{-x^2+2x+3}{(x^2-2x+5)^2} = 0$ | M1 |
$-x^2 + 2x + 3 = 0, -(x+1)(x-3) = 0$ |
$x = -1, 3$ | M1 A2 |
| (12) |

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5. (a) Show that $( 2 x + 3 )$ is a factor of $\left( 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 \right)$.\\
(b) Hence, simplify

$$\frac { 2 x ^ { 2 } + x - 3 } { 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 } .$$

(c) Find the coordinates of the stationary points of the curve with equation

$$y = \frac { 2 x ^ { 2 } + x - 3 } { 2 x ^ { 3 } - x ^ { 2 } + 4 x + 15 } .$$

\hfill \mbox{\textit{Edexcel C3  Q5 [12]}}

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Q1 7 Q2 9 Q3 10 Q4 10 Q5 12 Q6 13 Q7 14

Answer	Marks
let \(f(x) = 2x^3 - x^2 + 4x + 15\)	M1 A1
\(f(-\frac{3}{2}) = -\frac{27}{4} - \frac{9}{4} - 6 + 15 = 0\) ∴ \((2x+3)\) is a factor

Answer	Marks
\(\frac{dy}{dx} = \frac{1x(x^2-2x+5)-(x-1)(2x-2)}{(x^2-2x+5)^2} = \frac{-x^2+2x+3}{(x^2-2x+5)^2}\)	M1 A2
SP: \(\frac{-x^2+2x+3}{(x^2-2x+5)^2} = 0\)	M1
\(-x^2 + 2x + 3 = 0, -(x+1)(x-3) = 0\)
\(x = -1, 3\)	M1 A2
(12)