The polynomial $a{x}^3+b{x}^2+x-6$ has $(x+2)$ as a factor and leaves remainder $4$ when divided by $(x-2)$. Find the value of $a$ and $b$.

Let $P(x)=a{x}^3+b{x}^2+x-6$.Since $(x+2)$ is a factor, by the Factor Theorem, $P(-2)=0$:
$$a(-2{)}^3+b(-2{)}^2+(-2)-6=0$$
$$-8a+4b-8=0⟹-2a+b=2\text{— (Equation 1)}$$
Since $P(x)$ leaves a remainder of 4 when divided by $(x-2)$, by the Remainder Theorem, $P(2)=4$:
$$a(2{)}^3+b(2{)}^2+2-6=4$$
$$8a+4b-4=4⟹8a+4b=8⟹2a+b=2\text{— (Equation 2)}$$
Add Equation 1 and Equation 2:
$$(-2a+b)+(2a+b)=2+2$$
$$2b=4⟹b=2$$
Substitute $b=2$ back into Equation 2:
$$2a+2=2⟹2a=0⟹a=0$$
Thus, $a=0$ and $b=2$.