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 $a,{\text{ }}b \in \mathbb{R}$ .

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]

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.