Find factors of $x^{4} + 2 x^{3} - 7 x^{2} - 8 x + 12$

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Solution

$f (x) = X^{4} + 2 x^{3} - 7 x^{2} - 8 x + 12$
$A + x = 0$ $f (x) = 0$ By hit and trial
so, $(n - 1)$ is its factor
$f (x) = (x - 1) (x^{3} + 3 x^{2} - 4 x - 12)$
$= (x - 1) (x - 2) (x^{2} + 5 x + 6)$
$f (x) = (x - 1) (x - 2) (x + 5) (x + 7)$

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Similar questions

Using factor theorem,factorize the polynomial:

1) x⁴-2x³-7x²+8x+12

x^{4} - 2 x^{3} - 7 x^{2} + 8 x + 12

x^{4} + 2 x^{3} - 7 x^{2} - 8 x + 12 = 0

x^{4} - 2 x^{3} - 7 x^{2} + 8 x + 12

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