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Date November 2017 Marks available 3 Reference code 17N.1.AHL.TZ0.H_3
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Hence or otherwise Question number H_3 Adapted from N/A

Question

Consider the polynomial q ( x ) = 3 x 3 11 x 2 + k x + 8 .

Given that q ( x ) has a factor ( x 4 ) , find the value of k .

[3]
a.

Hence or otherwise, factorize q ( x ) as a product of linear factors.

[3]
b.

Markscheme

q ( 4 ) = 0     (M1)

192 176 + 4 k + 8 = 0   ( 24 + 4 k = 0 )     A1

k = 6     A1

[3 marks]

a.

3 x 3 11 x 2 6 x + 8 = ( x 4 ) ( 3 x 2 + p x 2 )

equate coefficients of x 2 :     (M1)

12 + p = 11

p = 1

( x 4 ) ( 3 x 2 + x 2 )     (A1)

( x 4 ) ( 3 x 2 ) ( x + 1 )     A1

 

Note:     Allow part (b) marks if any of this work is seen in part (a).

 

Note:     Allow equivalent methods (eg, synthetic division) for the M marks in each part.

 

[3 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 2—Functions » AHL 2.12—Factor and remainder theorems, sum and product of roots
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Topic 2—Functions

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