June 2022 Paper 1 Q11

11 The polynomial \(\mathrm{p}(x)\) is given by

\[\mathrm{p}(x) = x^3 + (b + 2)x^2 + 2(b + 2)x + 8\]

where \(b\) is a constant.

(a) Use the factor theorem to prove that \((x + 2)\) is a factor of \(\mathrm{p}(x)\) for all values of \(b\). [3 marks]

(b) The graph of \(y = \mathrm{p}(x)\) meets the \(x\)-axis at exactly two points.

(i) Sketch a possible graph of \(y = \mathrm{p}(x)\) [3 marks]

(ii) Given \(\mathrm{p}(x)\) can be written as\[\mathrm{p}(x) = (x + 2)(x^2 + bx + 4)\]find the value of \(b\).

Fully justify your answer. [4 marks]

Mark scheme (a)

Scheme	Marks	AO
Substitutes \(x = -2\) into \(\mathrm{p}(x)\) Condone missing brackets	M1	1.1a
Demonstrates clearly that \(\mathrm{p}(-2) = 0\) Must see numerical evaluation of powers of \(-2\) and either \(-8 + 4b + 8 - 4b - 8 + 8 = 0\) or \(-8 + 4(b + 2) - 4(b + 2) + 8 = 0\) or \(-8 + (b + 2)(4 - 4) + 8 = 0\)	A1	2.1
Concludes and states that \((x + 2)\) is a factor for all/any values of \(b\)	R1	2.4
	(3)

\[\mathrm{p}(-2) = (-2)^3 + (b + 2)(-2)^2 + 2(b + 2)(-2) + 8\]\[= -8 + 4b + 8 - 4b - 8 + 8\]\[= 0\]

Hence \((x + 2)\) is a factor of \(\mathrm{p}(x)\) for all values of \(b\)

Scheme	Marks	AO
(i) Sketches cubic graph with correct orientation and two turning points	B1	1.2
Sketches any cubic that would only ever meet the \(x\)-axis at exactly two points	M1	2.2a
Sketches a correctly orientated cubic graph that has a single root labelled \(x = -2\) \(y\) intercept labelled at 8 repeated root on the positive \(x\)-axis Ignore any value shown at the other root	A1	1.1b
	(3)
(ii) Deduces a pair of possible factors of \((x^2 + bx + 4)\) Either \((x + 2)^2\) or \((x - 2)^2\) or \((x + 1)(x + 4)\) or \((x - 1)(x - 4)\) or Uses \(b^2 - 4ac\) OE PI \(b = 4\) or \(b = -4\) or \(b^2 - 16\) seen	M1	2.2a
Identifies the quadratic factor as \((x - 2)^2\) or Obtains \(b^2 - 16 = 0\) OE PI by \(b = \pm 4\) or \(b = -4\)	R1	2.1
Obtains either \(b = \pm 4\) or \(b = -4\)	M1	1.1a
Rejects \(b = 4\) giving a reason and concludes \(b = -4\) Valid reasons would be: only one factor or root only one point of intersection with \(x\)-axis it would be \((x + 2)^3\) which only has one point of intersection	R1	2.4
	(4)
	(10 marks)