Date | November 2018 | Marks available | 3 | Reference code | 18N.2.AHL.TZ0.H_6 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Find | Question number | H_6 | Adapted from | N/A |
Question
Let P(x)=2x4−15x3+ax2+bx+c, where a, b, c∈R
Given that (x−5) is a factor of P(x), find a relationship between a, b and c.
Given that (x−5)2 is a factor of P(x), write down the value of P′(5).
Given that (x−5)2 is a factor of P(x), and that a=2, find the values of b and c.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
attempt to substitute x=5 and set equal to zero, or use of long / synthetic division (M1)
2×54−15×53+a×52+5b+c=0 A1
(⇒25a+5b+c=625)
[2 marks]
0 A1
[1 mark]
EITHER
attempt to solve P′(5)=0 (M1)
⇒8×53−45×52+4×5+b=0
OR
(x2−10x+25)(2x2+αx+β)=2x4−15x3+2x2+bx+c (M1)
comparing coefficients gives α = 5, β = 2
THEN
b = 105 A1
∴c=625−25×2−525
c = 50 A1
[3 marks]