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)=3x3−11x2+kx+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+4k+8=0 (24+4k=0) A1
k=−6 A1
[3 marks]
a.
3x3−11x2−6x+8=(x−4)(3x2+px−2)
equate coefficients of x2: (M1)
−12+p=−11
p=1
(x−4)(3x2+x−2) (A1)
(x−4)(3x−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.