Solve of $x : (x - 2) (x - 4) (x + 3) (x + 5) = 240$

3

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5

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6

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Solution

The correct option is A $5$
$(x - 2) (x - 4) (x + 3) (x + 5) = 240$
$\Rightarrow (x^{2} + x - 6) (x^{2} + x - 20) = 240$
$\Rightarrow (t - 6) (t - 20) = 240$
$\Rightarrow t^{2} - 26 t - 120 = 0$
$\Rightarrow (t - 30) (t + 4) = 0 \Rightarrow t = 30, - 4$
$\Rightarrow x^{2} + x = 30 \Rightarrow x = - 6, 5$
& $x^{2} + x = - 4 \Rightarrow x^{2} + x + 4 = 0$
As $D = (\sqrt{1 - 16}) = - 15 < 0$
$∴$ no real 'x'
hence $x = - 6, 5$

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\int \frac{x^{6} - 2 x^{4} + 3 x^{3} - 9 x^{2} + 4}{x^{5} - 5 x^{3} + 4 x} d x

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