Date | November 2019 | Marks available | 6 | Reference code | 19N.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(z)=az3−37z2+66z−10, where z∈C and a∈Z.
One of the roots of P(z)=0 is 3+i. Find the value of a.
Markscheme
METHOD 1
one other root is 3−i A1
let third root be α (M1)
considering sum or product of roots (M1)
sum of roots =6+α=37a A1
product of roots =10α=10a A1
hence a=6 A1
METHOD 2
one other root is 3−i A1
quadratic factor will be z2−6z+10 (M1)A1
P(z)=az3−37z2+66z−10=(z2−6z+10)(az−1) M1
comparing coefficients (M1)
hence a=6 A1
METHOD 3
substitute 3+i into P(z) (M1)
a(18+26i)−37(8+6i)+66(3+i)−10=0 (M1)A1
equating real or imaginary parts or dividing M1
18a−296+198−10=0 or 26a−222+66=0 or 10−66(3+i)+37(8+6i)18+26i A1
hence a=6 A1
[6 marks]