Question
Find the remainder when $2x^3 – 3x^2 + 4x + 7$ is divided by $2x + 1$
✓
Answer
Let $2 x+1=0$,
then $2 x-1$
$
\Rightarrow x =-\frac{1}{2}
$
Now substituting the value of $x$ in $f(x)$
$
\begin{aligned}
& f\left(-\frac{1}{2}\right)=2\left(-\frac{1}{2}\right)^2-3\left(-\frac{1}{2}\right)^2+4\left(-\frac{1}{2}\right)+7 \\
& =2\left(-\frac{1}{8}\right)-3\left(\frac{1}{4}\right)+4\left(-\frac{1}{2}\right)+7 \\
& =-\frac{1}{4}-\frac{3}{4}-2+7 \\
& =-1+-2+7 \\
& =4 \\
& \therefore \text { Remainder }=4 .
\end{aligned}
$
then $2 x-1$
$
\Rightarrow x =-\frac{1}{2}
$
Now substituting the value of $x$ in $f(x)$
$
\begin{aligned}
& f\left(-\frac{1}{2}\right)=2\left(-\frac{1}{2}\right)^2-3\left(-\frac{1}{2}\right)^2+4\left(-\frac{1}{2}\right)+7 \\
& =2\left(-\frac{1}{8}\right)-3\left(\frac{1}{4}\right)+4\left(-\frac{1}{2}\right)+7 \\
& =-\frac{1}{4}-\frac{3}{4}-2+7 \\
& =-1+-2+7 \\
& =4 \\
& \therefore \text { Remainder }=4 .
\end{aligned}
$
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