Question

If $x^3-6x^2+ax+b$ is exactly divisible by $x^2-3x+2$, then $12a+22b=?$

• A. 132
• B. 0.0
• C. 264
• D. - 66

Answer 1

The correct option is B 0.0

Given, $f\left(x\right)=x^3-6x^2+ax+b$ is exactly divisible by
$g\left(x\right)=x^2-3x+2$.

Let the remainder be $r\left(x\right)$.
Then $r\left(x\right)=0$

$x-3x^2-3x+2\left)\overline{x^3-6x^2+ax+b}\underset{–}{{}_{-}{x}^3\mp 3x^2\pm 2x}\underset{–}{-3x^2+(a-2)x+b\mp 3x^2\pm 9x\mp 6}\right(a-11)x+b+6$

$\therefore r\left(x\right)=(a-11)x+(b+6)$
$(a-11)x+(b+6)=0$

The polynomial will be zero if
$a-11=0\&b+6=0$
$\therefore a=11\&b=-6$

$\therefore 12a+22b=12\times 11+22\times (-6)=0$