Using factor theorem, show that g(x) is a factor of p(x), when
$p (x) = 2 x^{4} + x^{3} - 8 x^{2} - x + 6, g (x) = 2 x - 3$

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Solution

$p (x) = 2 x^{4} + x^{3} - 8 x^{2} - x + 6, g (x) = 2 x - 3 x = \frac{3}{2}$

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

$p (\frac{3}{2}) = 2 (\frac{3}{2})^{4} + (\frac{3}{2})^{3} - 8 (\frac{3}{2})^{2} - \frac{3}{2} + 6 = \frac{81}{8} + \frac{27}{8} - 18 - \frac{3}{2} + 6 = \frac{108}{8} - 12 - \frac{3}{2} = \frac{108 - 96 - 12}{8} = 0$

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

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Q. Using factor theorem, show that g(x) is a factor of p(x), when
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Q. Question 2
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

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Q. In the following case, use factor theorem to find whether

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Using factor theorem, show that g(x) is a factor of p(x), when p(x)=2x4+x3−8x2−x+6,g(x)=2x−3

p(x)=2x4+x3−8x2−x+6,g(x)=2x−3x=32 If p(x) is a multiple of g(x), the remainder will be zero. p(32)=2(32)4+(32)3−8(32)2−32+6=818+278−18−32+6=1088−12−32=108−96−128=0 Therefore p(x) is a multiple of g(x)

Using factor theorem, show that g(x) is a factor of p(x), when
$p (x) = 2 x^{4} + x^{3} - 8 x^{2} - x + 6, g (x) = 2 x - 3$