Date | May 2008 | Marks available | 6 | Reference code | 08M.1.hl.TZ2.2 |
Level | HL only | Paper | 1 | Time zone | TZ2 |
Command term | Find | Question number | 2 | Adapted from | N/A |
Question
The polynomial \(P(x) = {x^3} + a{x^2} + bx + 2\) is divisible by (x +1) and by (x − 2) .
Find the value of a and of b, where \(a,{\text{ }}b \in \mathbb{R}\) .
Markscheme
METHOD 1
As (x +1) is a factor of P(x), then P(−1) = 0 (M1)
\( \Rightarrow a - b + 1 = 0\) (or equivalent) A1
As (x − 2) is a factor of P(x), then P(2) = 0 (M1)
\( \Rightarrow 4a + 2b + 10 = 0\) (or equivalent) A1
Attempting to solve for a and b M1
a = −2 and b = −1 A1 N1
[6 marks]
METHOD 2
By inspection third factor must be x −1. (M1)A1
\((x + 1)(x - 2)(x - 1) = {x^3} - 2{x^2} - x + 2\) (M1)A1
Equating coefficients a = −2, b = −1 (M1)A1 N1
[6 marks]
METHOD 3
Considering \(\frac{{P(x)}}{{{x^2} - x - 2}}\) or equivalent (M1)
\(\frac{{P(x)}}{{{x^2} - x - 2}} = (x + a + 1) + \frac{{(a + b + 3)x + 2(a + 2)}}{{{x^2} - x - 2}}\) A1A1
Recognising that \((a + b + 3)x + 2(a + 2) = 0\) (M1)
Attempting to solve for a and b M1
a = −2 and b = −1 A1 N1
[6 marks]
Examiners report
Most candidates successfully answered this question. The majority used the factor theorem, but a few employed polynomial division or a method based on inspection to determine the third linear factor.