Question

If the polynomial $f\left(x\right)=a{x}^3+bx-c$ is divisible by the polynomial $g\left(x\right)=x^2+bx+c$, then $ab=?$

• A. $-\frac{1}{c}$
• B. -1
• C. 1
• D. $\frac{1}{c}$

Answer 1

The correct option is C 1

$a{x}^3+bx-c$ is divisible by $x^2+bx+c\Rightarrow$ remainder = 0

i.e. $f\left(x\right)=g\left(x\right)q\left(x\right)+r\left(x\right)$
where, $r\left(x\right)=0$

$ax-ab{x}^2+bx+c)\overline{a{x}^3+bx-c}\underset{–}{{}_{-}a{x}^3\pm ab{x}^2\pm acx}\underset{–}{-ab{x}^2+(b-ac)x-c\mp ab{x}^2\mp a{b}^{2}x\mp abc}a{b}^{2}x+(b-ac)x+abc-c$

So, $r\left(x\right)=(a{b}^2+b-ac)x+(abc-c)=0$

The $(a{b}^2+b-ac)x+(abc-c)$ will be zero,
If $a{b}^2+b-ac=0$ and $abc-c=0$

$a{b}^2+b-ac=0$ is a quadratic polynomial, lets take other equation.

$c(ab-1)=0$
$\Rightarrow c=0;ab-1=0\therefore ab=1$