Using factor theorem, show that g(x) is a factor of p(x), when
$p (x) = 3 x^{3} + x^{2} - 20 x + 12, g (x) = 3 x - 2$

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Solution

$p (x) = 3 x^{3} + x^{2} - 20 x + 12 g (x) = 3 x - 2 x = \frac{2}{3}$

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

$p (\frac{2}{3}) = 3 (\frac{2}{3})^{3} + (\frac{2}{3})^{2} - 20 (\frac{2}{3}) + 12 = \frac{8}{9} + \frac{4}{9} - \frac{40}{3} + 12 = \frac{12}{9} - \frac{40}{3} + 12 = \frac{4}{3} - \frac{40}{3} + \frac{36}{3} = \frac{4 - 40 + 36}{3} = 0$

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

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