Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
$(p x) = x^{3} - 6 x^{2} + 2 x - 4, g (x) = 1 - \frac{3}{2} x$

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Solution

$p (x) = x^{3} - 6 x^{2} + 2 x - 4 g (x) = 1 - \frac{3}{2} x 1 - \frac{3}{2} x = 0 \frac{3 x}{2} = 1$

By remainder theorem, when p(x) is divided by ( 3x+2), then the remainder = $p (\frac{- 3}{2}) 3 x = 2 x = \frac{2}{3}$ .

Putting x = $\frac{2}{3}$ in p(x), we get

$p (\frac{2}{3}) = (\frac{2}{3})^{3} - 6 (\frac{2}{3})^{2} + 2 (\frac{2}{3}) - 4 = \frac{8}{27} - \frac{24}{9} + \frac{4}{3} - 4 = \frac{8}{27} - \frac{72}{27} + \frac{36}{27} - \frac{108}{27} = \frac{8 - 72 + 36 - 108}{27} = \frac{- 136}{27}$

∴ Remainder = $\frac{- 136}{27}$

Thus, the remainder when p(x) is divided by g(x) is $\frac{- 136}{27}$ .

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Using the remainder theorem, find the remainder, when p(x) is divided by g(x), where
$(p x) = x^{3} - a x^{2} + 6 x - a, g (x) = x - a$

p (x)

g (x)

p (x) = x^{3} - 2 x^{2} - 4 x - 1

g (x) = x + 1

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