Question

If the polynomial $a{x}^3+b{x}^2-c$ is divisible by $x^2+bx+c$ then $ab$ is equal to?

Answer 1

Step 1: Divide $a{x}^3+b{x}^2-c$ by $x^2+bx+c$

It is given that the polynomial $a{x}^3+b{x}^2-c$ is divisible by $x^2+bx+c$.

Let, $f\left(x\right)=a{x}^3+b{x}^2-c$

And, $g\left(x\right)=x^2+bx+c$

On dividing $f\left(x\right)$ by $g\left(x\right)$, we see,

$x^2+bx+c\begin{array}{cc}ax& -ab\end{array}\begin{array}{lcl}a{x}^3& +bx& -c\end{array}\begin{array}{ccl}a{x}^3& +ab{x}^2& +acx\\ -& -& -\end{array}\begin{array}{cccr}& -ab{x}^2& +\left(b-ac\right)x& -c\end{array}\begin{array}{crcc}& -ab{x}^2& -a{b}^{2}x& -abc\end{array}\begin{array}{cccc}& +& +& +\end{array}\begin{array}{ccc}\left(b-ac+a{b}^2\right)x& +abc& -c\end{array}$

Here, $\left(b-ac+a{b}^2\right)x+abc-c$ is the remainder.

Step 2: Calculate the value of $ab$

Now, according to the question, the polynomial $a{x}^3+b{x}^2-c$ is divisible by $x^2+bx+c$, i.e., the remainder is zero.

So, equating the remainder to zero, we get,

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

$\Rightarrow$ $\left(b-ac+a{b}^2\right)x+\left(ab-1\right)c=0$

Now, for the remainder to be zero, it is necessary that both the terms in the above equation are zero.

i.e., $\left(b-ac+a{b}^2\right)x=0$ and $\left(ab-1\right)c=0$

Taking, $\left(ab-1\right)c=0$

$\Rightarrow$ $\left(ab-1\right)=0$

$\Rightarrow$ $ab-1=0$

$\Rightarrow$ $ab=1$

Hence, the required value of $ab$ is $1$.