$4 x^{3} - 8 x^{2} + 8 x + 1,$ when divided by a polynomial $g (x)$ gives $2 x - 1$ as quotient and $x - 3$ as remainder. Find $g (x) .$

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Solution

Let $f (x) = 4 x^{3} - 8 x^{2} + 8 x + 1$

$q (x) = 2 x - 1$

$r (x) = x + 3$

$f (x) = g (x) \times q (x) + r (x)$

Degree of $g (x)$ be 2. as degree of $f (x)$ is 3.
$∴ g (x) = a x^{2} + b x + c$

$4 x^{3} - 8 x^{2} + 8 x + 1 = (a x^{2} + b x + c) (2 x - 1) + x + 3$

$= 2 a x^{3} + 2 b x^{2} + 2 c x - a x^{2} - b x + c + x + 3$

$= 2 a x^{3} + (2 b - a) x^{2} + (2 c - b + 1) + (3 - c)$

Equating coefficient of power of $x,$

$2 a = 4 \Rightarrow a = \frac{4}{2} = 2$

$2 b - a = - 8 \Rightarrow b = - 3$

$2 c - b + 1 = 8 \Rightarrow 2 c + 3 = 7 \Rightarrow c = 2$
$∴ g (x) = 2 x^{2} - 3 x + 2$

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