$If x^{2} - 4 x - 12 = (x + m) (x + n)$ ,
then find the value of $(m + n) .$

2

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- 2

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4

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- 4

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Solution

The correct option is D $- 4$
Given:
$x^{2} - 4 x - 12 = (x + m) (x + n) . . . (i)$

Now, comparing the expression $x^{2} - 4 x - 12$ with the identity
$x^{2} + (a + b) x + a b,$
we note that,
$(a + b) = - 4 a n d a b = - 12$
So,
$(- 6) + 2 = - 4 a n d (- 6) (2) = - 12$

Hence,
$x^{2} - 4 x - 12$
$= x^{2} - 6 x + 2 x - 12$
$= x (x - 6) + 2 (x - 6)$
$= (x - 6) (x + 2)$

Now, from (i)
$x^{2} - 4 x - 12 = (x + m) (x + n)$
$\Rightarrow (x + m) (x + n) = (x - 6) (x + 2)$
After comparing the above equation, we get
$m = - 6 and n = 2$

Therefore,
$(m + n) = - 6 + 2 = - 4$

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