Use the Factor Theorem to determine whether $g\left(x\right)$ is a factor of $p\left(x\right)$ in each of the following cases:

(i) $p\left(x\right)=2x^3+x^2-2x-1, g\left(x\right)=x+1$

(ii) $p\left(x\right)=x^3+3x^2+3x+1, g\left(x\right)=x+2$

(iii) $p\left(x\right)=x^3-4x^2+x+6, g\left(x\right)=x-3$

(i) Using factor theorem, we get:

For $g\left(x\right)=0$, the value of $x$ is given by:

$\begin{array}{rcl}x+1& =& 0\\ x& =& -1\end{array}$

Now,

$\begin{array}{rcl}p\left(x\right)& =& 2x^3+x^2-2x-1\\ p\left(-1\right)& =& 2{\left(-1\right)}^3+{\left(-1\right)}^2-2\left(-1\right)-1\\ & =& -2+1+2-1\\ & =& 0\end{array}$

Hence, $g\left(x\right)$ is a factor of $p\left(x\right)$

(ii) Using factor theorem, we get:

For $g\left(x\right)=0$, the value of $x$ is given by:

$\begin{array}{rcl}x+2& =& 0\\ x& =& -2\end{array}$

Now,

$\begin{array}{rcl}p\left(x\right)& =& x^3+3x^2+3x+1\\ p\left(-2\right)& =& {\left(-2\right)}^3+3{\left(-2\right)}^2+3\left(-2\right)+1\\ & =& -8+12-6+1\\ & =& -1\ne 0\end{array}$

Hence, $g\left(x\right)$ is not a factor of $p\left(x\right)$