Question:
When $p(x)=4 x^{3}-12 x^{2}+11 x-5$ is divided by $(2 x-1)$, the remainder is
(a) 0
(b) $-5$
(c) $-2$
(d) 2
Solution:
(c) −2
$2 x-1=0 \Rightarrow x=\frac{1}{2}$
By the remainder theorem, we know that when $p(x)$ is divided by $(2 x-1)$, the remainder is $p\left(\frac{1}{2}\right)$.
Now, we have:
$p\left(\frac{1}{2}\right)=4 \times\left(\frac{1}{2}\right)^{3}-12 \times\left(\frac{1}{2}\right)^{2}+11 \times \frac{1}{2}-5$
$=\frac{1}{2}-3+\frac{11}{2}-5$
$=-2$