The number of real roots of the equation $(x + 1) (x + 2) (x + 3) (x + 4) = 120$ is

4

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3

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2

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0

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Solution

The correct option is C $2$
Given equation
$(x + 1) (x + 2) (x + 3) (x + 4) = 120$
$\Rightarrow (x + 1) (x + 4) (x + 2) (x + 3) = 120 \Rightarrow (x^{2} + 5 x + 4) (x^{2} + 5 x + 6) = 120$

Assuming $x^{2} + 5 x = t$
$\Rightarrow (t + 4) (t + 6) = 120 \Rightarrow t^{2} + 10 t - 96 = 0 \Rightarrow (t + 16) (t - 6) = 0 \Rightarrow t = 6 or t = - 16$
Now,
$x^{2} + 5 x = - 16 \Rightarrow x^{2} + 5 x + 16 = 0 \Rightarrow D = 25 - 64 < 0$
Non real roots.

$x^{2} + 5 x - 6 = 0 \Rightarrow (x + 6) (x - 1) = 0 \Rightarrow x = - 6, 1$

Hence, the number of real roots are $2$ .

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