Question

I'm not sure how to find zeros and multipilicities

Answer 1

If $p(x)$ is a polynomial, then the zeros of $p(x)$ are the values of $x$ that satisfy $p(x)=0$.

So for (1), the zeros of $f(x)=(x-4)(x-1)(x+3)$ are the solutions to

$(x-4)(x-1)(x+3)=0\implies x=4,x=1,x=-3$

You can think of the multiplicity of each root as the number of times a root satisfies this equation. In this case, the multiplicity of each root is 1, because there is only one factor of $x-4$ or $x-1$ or $x+3$.

For (2), the zeros of $g(x)$ are the solutions to

$(x+1)(x-2)^2(x-4)=0\implies x=-1,x=2,x=4$

and the multiplicity of the zero $x=2$ is 2 because there are two factors of $x-2$ in $g(x)$. The multiplicity of the other roots would be 1.