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.