Solve the equation $x^{4} + x^{3} - 16 x^{2} - 4 x + 48 = 0$ , if the product of two its roots is $8$ .

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Solution

$∴$ Resulting equation is $x^{3} + 3 x^{2} - 10 x - 24$
$f (- 2)$ is the other root
Resulting equation is $x^{2} + x - 12 = 0$
$x^{2} + 4 x - 3 x - 12 = 0$
$\Rightarrow x (x + 4) - 3 (x + 4) = 0$
$\Rightarrow (x + 4) (x - 3) = 0$
$∴ x = - 4, 3$
Hence the roots are $- 4, - 2, 2, 3$

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