Question

If $a-b=3$, $a+b+x=2$, then the value of $(a-b)[x^3-2a{x}^2+a^{2}x-(a+b)b^2]$ is

• A. $84$
• B. $48$
• C. $32$
• D. $36$

Answer 1

Answer: (B)

$48$

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

We know that the remainder theorem states the following: If you divide a polynomial $f\left(x\right)$ by $(x-h)$, then the remainder is $f\left(h\right)$.

By remainder theorem, since $a-b=3$, therefore, $(x-3)$ is a factor of $f\left(x\right)$.

Thus, $x=3$ is one of the roots,

Using this value in $a+b+x=2$, we get,

$a+b+3=2\Rightarrow a+b=2-3\Rightarrow a+b=-1$

Add the given equation $a-b=3$ in $a+b=-1$ as follows:

$a-b+a+b=3-1\Rightarrow 2a=2\Rightarrow a=\frac{2}{2}\Rightarrow a=1$

Substitute the value of $a$ in $a-b=3$ as follows:

$1-b=3\Rightarrow b=1-3\Rightarrow b=-2$

Therefore, $a=1,b=-2$ and also $x=3$.

Putting all the three values in $f\left(x\right)$, we have:

$f\left(x\right)=(1-(-2\left)\right)[3^3-2\times 1\times 3^2+1^2\times 3-(1-2\left)\right(-2{)}^2]=(1+2\left)\right[27-18+3-(-4)]=3(27-18+3+4)=3(34-18)=3\times 16=48$

Hence, the value of $(a-b)[x^3-2a{x}^2+a^{2}x-(a+b\left)b^2\right]=48$.