Find the remainder when $f (x) = x^{3} - 6 x^{2} + 2 x - 4$ divided by g(x) = 1 - 2x

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Solution

f(x)= $x^{3} - 6 x^{2} + 2 x - 4$
Let 1 - 2x = 0 $\Rightarrow$ 2x = 1 $\Rightarrow x = \frac{1}{2}$
Remainder = $\Rightarrow f (x) = f (\frac{1}{2}) = {(\frac{1}{2})}^{3} - 6 {(\frac{1}{2})}^{2} + 2 (\frac{1}{2}) - 4$
$= \frac{1}{8} - \frac{3}{2} + 1 - 4 = \frac{1 - 12 + 8 - 32}{8} = - \frac{35}{8}$

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