Question

$\text{The polynomials}a{x}^3+4x^2+3x–4\text{and}{x}^3–4x+a\text{leave the same remainder}\text{when divided by}(x–3).\text{Find the value}\text{of a}.$

• A. $-1$
• B. $-2$
• C. $-3$
• D. $-4$

Answer 1

The correct option is A $-1$

$\text{Given polynomials:}p\left(x\right)=a{x}^3+4x^2+3x-4q\left(x\right)=x^3-4x+a$

$\text{Using remainder theorem, the}\text{remainders when}p\left(x\right)\text{and}q\left(x\right)\text{are divided by}(x-3)\text{are}p\left(3\right)\text{and}q\left(3\right)\text{respectively.}$

$p\left(3\right)=a\times {\left(3\right)}^3+4\times {\left(3\right)}^2+3\times 3-4p\left(3\right)=27a+41$

$q\left(3\right)={\left(3\right)}^3-4\times 3+a$
$q\left(3\right)=15+a$

$\text{Given: The remainder is the same.}\Rightarrow p\left(3\right)=q\left(3\right)\Rightarrow 27a+41=15+a\Rightarrow 26a=-26\Rightarrow a=-1$