Using factor theorem, show that g(x) is a factor of p(x), when
$p (x) = x^{4} - x^{2} - 12, g (x) = x + 2$

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Solution

$p (x) = x^{4} - x^{2} - 12 g (x) = x + 2 x = - 2$

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

$p (- 2) = (- 2)^{4} - (- 2)^{2} - 12 = 16 - 4 - 12 = 0$

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

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Q. Question 2
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Using factor theorem, show that g(x) is a factor of p(x), when p(x)=x4−x2−12,g(x)=x+2

p(x)=x4−x2−12g(x)=x+2x=−2 If p(x) is a multiple of g(x), the remainder will be zero. p(−2)=(−2)4−(−2)2−12=16−4−12=0 Therefore p(x) is a multiple of g(x)

Using factor theorem, show that g(x) is a factor of p(x), when
$p (x) = x^{4} - x^{2} - 12, g (x) = x + 2$

$p (x) = x^{4} - x^{2} - 12 g (x) = x + 2 x = - 2$

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

$p (- 2) = (- 2)^{4} - (- 2)^{2} - 12 = 16 - 4 - 12 = 0$

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