Question

The remainder if $a{x}^3+b{x}^2+cx+d$ is divided by $ax+b$

• A. $ad-bc$
• B. $\frac{1}{a}(ad-bc)$
• C. $\frac{a-bc}{d}$
• D. $\frac{a+b+cd}{2}$

Answer 1

Answer: (B)

$\frac{1}{a}(ad-bc)$

Here, $p\left(x\right)=a{x}^3+b{x}^2+cx+d$ and factor of $ax+b$ is
$ax+b=0$
$x=-\frac{b}{a}$
$p(-\frac{b}{a})=a(-\frac{b}{a}{)}^3+b(-\frac{b}{a}{)}^2+c(-\frac{b}{a})+d$
$p(-\frac{b}{a})=-\frac{b^3}{a^2}+\frac{b^3}{a^2}-\frac{bc}{a}+d$
$p(-\frac{b}{a})=-\frac{bc}{a}+d$
$p(-\frac{b}{a})=\frac{1}{a}(ad-bc)$
when $a{x}^3+b{x}^2+cx+d$ is divided by $ax+b$ then remainder is $p(-\frac{b}{a})=\frac{1}{a}(ad-bc)$
Option B is correct.