Find the remainder when $4 x^{2} - 8 x + 3$ is divided by $2 x - 1$ .

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Solution

As per the remainder theorem
If a polynomial $P (x)$ is divided by $x - a,$ the remainder obtained is $P (a)$
As per the question, we have $2 x - 1$
to find $a,$ put $2 x - 1 = 0$
$⟹ x = \frac{1}{2}$
so, put $x = \frac{1}{2}$ in $P (x) = 4 x^{2} - 8 x + 3$
$⟹ P (\frac{1}{2}) = 4 \times {(\frac{1}{2})}^{2} - 8 (\frac{1}{2}) + 3$
$⟹ P (\frac{1}{2}) = 1 - 4 + 3 = 0$
Hence Remainder is $0$

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