Question

Check whether $p\left(x\right)$ is multiple of $g\left(x\right)$ or not
(i) $p\left(x\right)=x^3+27x^2-8x+18,$ $g\left(x\right)=x-1$
(ii) $p\left(x\right)=2\sqrt{2}{x}^3-5\sqrt{2}{x}^2+7\sqrt{2},g\left(x\right)=x+1$

Answer 1

Part (1),

Consider the given polynomial $p\left(x\right)=x^3+27x^2-8x+18$ and $g\left(x\right)=x-1$

Let put $g\left(x\right)=x-1$ then

$x-1=0$

$x=1$

Put and get value of

$p\left(x\right)={\left(1\right)}^3+27{\left(1\right)}^2-8\left(1\right)+18$

$p\left(x\right)=1+27-8+18$

$p\left(x\right)=38\ne 0$

Because $p\left(x\right)\ne 0$

Hence, $p\left(x\right)$ is not multiple of $g\left(x\right)$.

Part (2),

Consider the given polynomial $p\left(x\right)=2\sqrt{2}{x}^3-5\sqrt{2}{x}^2+7\sqrt{2}$ and $g\left(x\right)=x+1$

Let put $g\left(x\right)=x+1$ then

$x+1=0$

$x=-1$

Put and get value of

$p\left(x\right)=2\sqrt{2}{(-1)}^3-5\sqrt{2}{(-1)}^2+7\sqrt{2}$

$p\left(x\right)=-2\sqrt{2}+5\sqrt{2}+7\sqrt{2}$

$p\left(x\right)=10\sqrt{2}\ne 0$

Because $p\left(x\right)\ne 0$

Hence, $p\left(x\right)$ is not multiple of $g\left(x\right)$.

Hence, this is the complete solution.