By actual division, show that $x^{2} - 3$ is a factor of $2 x^{4} + 3 x^{3} - 2 x^{2} - 9 x - 12$

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Solution

Let $f (x) = 2 x^{4} + 3 x^{3} - 2 x^{2} - 9 x - 12$ and $g (x) = x^{2} - 3$

Quotient $q (x) = 2 x^{2} + 3 x + 4$
Remainder $r (x) = 0$
Since, the remainder is 0.
Hence, $x^{2} - 3$ is a factor of $2 x^{4} + 3 x^{3} - 2 x^{2} - 9 x - 12$

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