Question
If $p(x) = 4x^3 - 3x^2 + 2x - 4$ find the remainderwhen $p(x)$ is divided by:
$x+\frac{1}{2}$.
$x+\frac{1}{2}$.
✓
Answer
$p(x)=4 x^3-3 x^2+2 x-4$ .... (i)
By the remainder theorem the required remainder
$= p \left(-\frac{1}{2}\right) \text {. }$
Put $x=\left(-\frac{1}{2}\right)$ in equation (i) we get
$ p\left(-\frac{1}{2}\right)=4\left(-\frac{1}{2}\right)^3-3\left(-\frac{1}{2}\right)^2+2\left(-\frac{1}{2}\right)-4$
$=4 \times\left(\frac{1}{8}\right)-3 \times \frac{1}{4}+2 \times\left(-\frac{1}{2}\right)-4$
$=-\frac{1}{2}-\frac{3}{4}-1-4$
$=\frac{-2-3-4-16}{4}$
$=-\frac{25}{4} $
Hence, the remainder is $-\frac{25}{4}$.
By the remainder theorem the required remainder
$= p \left(-\frac{1}{2}\right) \text {. }$
Put $x=\left(-\frac{1}{2}\right)$ in equation (i) we get
$ p\left(-\frac{1}{2}\right)=4\left(-\frac{1}{2}\right)^3-3\left(-\frac{1}{2}\right)^2+2\left(-\frac{1}{2}\right)-4$
$=4 \times\left(\frac{1}{8}\right)-3 \times \frac{1}{4}+2 \times\left(-\frac{1}{2}\right)-4$
$=-\frac{1}{2}-\frac{3}{4}-1-4$
$=\frac{-2-3-4-16}{4}$
$=-\frac{25}{4} $
Hence, the remainder is $-\frac{25}{4}$.
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