$x^{3} + a x^{2} - 7 x - 6$ can be factorised into 3 linear factor 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 A $a = 0$
Putting $a = 0$ , The given exp.
$= x^{3} - 7 x - 6$
$= x^{3} - x - 6 x - 6$
$= x (x^{2} - 1) - 6 (x + 1)$
$= (x + 1) [x (x - 1) - 6]$
$= (x + 1) [x^{2} - 3 x + 2 x - 6]$
$= (x + 1) [x (x - 3) + 2 (x - 3)]$
$= (x + 1) (x + 2) (x - 3)$
Hence, the given expression can be factorised into 3 linear factors, when $a = 0$

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