Question

$P\left(x\right)$ is a polynomial of degree 3 set.
$(x-1{)}^2$ is a factor of $P\left(x\right)+2$
$(x+1{)}^2$ is a factor of $P\left(x\right)-2$
Find $\to P\left(0\right)$

Answer 1

$P\left(x\right)=a{x}^3+b{x}^2+cx+d$
$P\left(x\right)+2=a{x}^3+b{x}^2+cx+d+2$

${(x-1)}^2$ is a factor of $P\left(x\right)+2$
i.e. $x=1\Rightarrow P\left(1\right)+2=a+b+c+d+2=0$ ........ $\left(i\right)$

Also, $P\left(x\right)-2=a{x}^3+b{x}^2+cx+d-2$

${(x+1)}^2$ is a factor of $P\left(x\right)-2$
i.e. $x=-1\Rightarrow P(-1)-2=-a+b-c+d-2=0$ ........ $\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$

$-b=d$
From $\left(i\right)$ and $\left(ii\right)$, we get

$c=-(a+2)$

Check the image

At $x=1$

$(c-a-2b+4a)+(2-2b+2a)=0$

$c+3a-2b+2-2b+2a=0$

$c+5a-4b+2=0$

$-a-2+5a-4b+2=0$ ........... $[\because c=-(a+2\left)\right]$

$4a-4b=0$

$a=b$

At $x=-1$

$(2b-2)(-1)-2=0$

$b=0$

$P\left(x\right)=b{x}^3+b{x}^2-(b+2)x-b$

$P\left(0\right)=-b=0$