# the remainder is

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:

$p\left(\frac{1}{2}\right)=4 \times\left(\frac{1}{2}\right)^{3}-12 \times\left(\frac{1}{2}\right)^{2}+11 \times \frac{1}{2}-5$

$=\frac{1}{2}-3+\frac{11}{2}-5$

$=-2$