$x^{3} + a x^{2} - 7 x - 6$ can be factorized into 3 linear factors when

a = 0

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a = 1

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a = 2

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a = 3

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Solution

The correct option is B $a = 0$
Given-
$f (x) = x^{3} + a x^{2} - 7 x - 6$
Option A $\to$ Putting $a = 0$
$f x = x^{3} - 7 x - 6$
$= x^{3} - 3 x^{2} + 3 x^{2} - 9 x + 2 x - 6$
$= x^{2} (x - 3) + 3 x (x - 3) + 2 (x - 3)$
$= (x - 3) (x^{2} + 3 x + 2)$
$= (x - 3) (x^{2} + 2 x + x + 2)$
$= (x - 3) (x + 1) (x + 2)$
So option A is correct option B, C and D are obviously not correct as the factorization of f(x) is complete in 3 simplest factors and no other factors are possible.

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