Question

If $(x-2)$ is a common factor of $x^3-4x^2+ax+b$ and $x^3-a{x}^2+bx+8$, then the values of $a$ and $b$ are respectively :

• A. $3$ and $5$
• B. $2$ and $-4$
• C. $4$ and $0$
• D. $0$ and $4$

Answer 1

Answer: (C)

$4$ and $0$

Let $f\left(x\right)=x^3-4x^2+ax+b$ and $g\left(x\right)=x^3-a{x}^2+bx+8$
By factor theorem, $f\left(2\right)$ and $g\left(2\right)$ both are equal to $0.$
Thus, $2^3-4\times 2^2+2a+b=0$ and $2^3-2^{2}a+2b+8=0$
$\therefore$ $8-16+2a+b=0$ $and$ $8-4a+2b+8=0$
$\therefore$ $2a+b=8$ $and$ $-2a+b=-8$
$\therefore$ adding both equations,

$2b=0$ $=>$ $b=0$
$\therefore$ $a=4.$