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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.

Syllabus sections

Topic 2 - Core: Functions and equations » 2.5 » The factor and remainder theorems.

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