Question

When $3x^3+2x^2-x+4$ is divided by $3+2x$ the remainder is $\frac{-1}{8}$.

• A. True
• B. False

Answer 1

The correct option is A

True

$P\left(x\right)=3x^3+2x^2-x+4$

If $P\left(x\right)$ is divided by $(x-a)$, then the remainder is $P\left(a\right)$

$2x+3=0\Rightarrow x=\frac{-3}{2}$

So, the remainder is the value of polynomial at $\frac{-3}{2}$

Substituting $x=\frac{-3}{2}$ in $p\left(x\right)$

$p\left(\frac{-3}{2}\right)=3(\frac{-3}{2}{)}^3+2(\frac{-3}{2}{)}^2-\left(\frac{-3}{2}\right)+4$

$p\left(\frac{-3}{2}\right)=3\left(\frac{-27}{8}\right)+2\left(\frac{9}{4}\right)-\left(\frac{-3}{2}\right)+4$

$p\left(\frac{-3}{2}\right)=\frac{-81}{8}+\frac{9}{2}+\frac{3}{2}+4$

$p\left(\frac{-3}{2}\right)=\frac{-81+36+12+32}{8}$

$p\left(\frac{-3}{2}\right)=\frac{-81+80}{8}$

$p\left(\frac{-3}{2}\right)=\frac{-1}{8}$

$\therefore$ The remainder is $\frac{-1}{8}.$

Thus, the given statement is true.