June 2025 Paper 2 Q4
4
(a) Use the factor theorem to show that \((x-3)\) is a factor of \(4x^3 - 8x^2 - 11x - 3\). [1]
(b) Hence show that \(4x^3 - 8x^2 - 11x - 3 = (x-3)(ax+b)^2\), where \(a\) and \(b\) are integers to be determined. [2]
(c) Sketch the graph of \(y = 4x^3 - 8x^2 - 11x - 3\) on the axes in the Printed Answer Booklet. [2]
| Scheme | Marks | AO |
|---|---|---|
| \(4\times 3^3 - 8\times 3^2 - 11\times 3 - 3\) \([= 108 - 72 - 33 - 3] = 0\) | B1 | 1.1 |
| [1] |
Notes
B1: arithmetic must be correct and 4 terms must be seen
| Scheme | Marks | AO |
|---|---|---|
| \((4x^2 + 4x + 1)\) seen | M1 | 2.1 |
| \((x-3)(2x+1)^2\) isw or \(a = 2,\ b = 1\) isw | A1 | 1.1 |
| [2] |
Notes
M1: may be embedded in long/synthetic division; allow one coefficient error
A1: allow \((x-3)(2x+1)(2x+1)\); mark the final answer; supporting working must be seen for the award of this mark
Alternative method
| Scheme | Marks |
|---|---|
| \(a^2 = 4,\ -3a^2 + 2ab = -8,\) \(b^2 - 6ab = -11,\ -3b^2 = -3\) | M1 |
| \(a = 2,\ b = 1\) isw | A1 |
M1: for any two equations from equating coefficients; may be embedded in expansion; allow one coefficient error
A1: supporting working must be seen for the award of this mark
| Scheme | Marks | AO |
|---|---|---|
 | M1 A1 | 1.1 1.1 |
| [2] |
Notes
M1: cubic of correct orientation; curve touches \(x\)-axis on correct side of origin for their \(-\frac{b}{a}\) and intersects positive \(x\)-axis;
A1: intercepts at \(\left(-\frac{1}{2},\ 0\right)\), \((3,\ 0)\) and \((0,\ -3)\) labelled or identified on sketch; local minimum in 4th quadrant