Question

(a) Check whether $(x-1)$ is a factor of the polynomial $p\left(x\right)=6x^3+3x^2$
(b) What first degree polynomial added to $p\left(x\right)$ gives a polynomial for which $(x^2-1)$ is a factor?

Answer 1

(a) According to the factor theorem, if $(x-a)$ is a factor of $f\left(x\right)$, then $f\left(a\right)=0$

Hence, $(x-1)$ will be a factor of $p\left(x\right)=6x^3+3x^2$ when $p\left(1\right)=0$
$p\left(1\right)=6(1{)}^3+3(1{)}^2=6+3=9\ne 0$
$\therefore p\left(1\right)\ne 0$

Hence, $(x-1)$ is not a factor of $p\left(x\right)=6x^3+3x^2$

(b) Let $f\left(x\right)$ be the required polynomial.
$p\left(x\right)=6x^3+3x^2$
Let the first degree polynomial to be added be $ax+b$.
$f\left(x\right)=p\left(x\right)+ax+b$
If $x^2-1$ is a factor of the $f\left(x\right)$, then $f(-1)=0$ and $f\left(1\right)=0$ $[x^2-1=(x+1\left)\right(x-1\left)\right]$
$\Rightarrow f(-1)=p(-1)+a(-1)+b=0$
$\Rightarrow 6(-1{)}^3+3(-1{)}^2-a+b=0$
$\Rightarrow -6+3-a+b=0$
$\Rightarrow a-b=-3....\left(i\right)$
Also, $f\left(1\right)=p\left(1\right)+a\left(1\right)+b=0$
$\Rightarrow 6(1{)}^3+3(1{)}^2+a+b=0$
$\Rightarrow a+b=-9....\left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right)$, we get:
$2a=-12$

$\therefore a=-6$

Substituting $a=-6$ in $\left(i\right)$, we get:

$b=-3$
$\therefore ax+b=-6x-3$

Hence, $-6x-3$ is the first degree polynomial that must be added to $p\left(x\right)$ to get a polynomial that has $(x^2-1)$ as a factor.