Using factor theorem, show that g(x) is a factor of p(x), when
$p (x) = 2 x^{4} + 9 x^{3} + 6 x^{2} - 11 x - 6, g (x) = x - 1$

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Solution

$p (x) = 2 x^{4} + 9 x^{3} + 6 x^{2} - 11 x - 6 g (x) = x - 1 x = 1$

If p(x) is a multiple of g(x), the remainder will be zero.

$p (1) = 2 (1)^{4} + 9 (1)^{3} + 6 (1)^{2} - 11 (1) - 6 = 2 + 9 + 6 - 11 - 6 = 0$

Therefore p(x) is a multiple of g(x)

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