$If (x - 1) and (x + 2) are the factors of the cubic polynomial$
$x^{3} + (3 a + 1) x^{2} + b x - 18, then a + b =$

[2 marks]

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Solution

The correct option is C 10
Given,
$Let f (x) = x^{3} + (3 a + 1) x^{2} + b x - 18$

$Since (x - 1) and (x + 2) are the factors of x^{3} + (3 a + 1) x^{2} + b x - 18$

$\Rightarrow x = 1 and x = - 2 satisfy f(1) = 0 and f(-2) = 0$

$f (1) = 0$
$\Rightarrow 1^{3} + (3 a + 1) + b - 18 = 0 \Rightarrow 3 a + b = 16 . . . (1) f (- 2) = 0 \Rightarrow (- 2)^{3} + (3 a + 1) (- 2)^{2} - 2 b - 18 = 0 \Rightarrow - 8 + 12 a + 4 - 2 b - 18 = 0 \Rightarrow 6 a - b = 11 . . . (2)$
On solving equations (1) and (2), we get,
$a = 3, b = 7$
$∴ a + b = 10$

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