Question

How do you find the zeros of `(-x+1)(x+2)(x-3)^2`?

Answer 1

There are four zero in total: at `x=-2, x=1, x=3 and x=3`

graph{(-x+1)(x+2)(x-3)^2 [-7.116, 10.05, -12.5, 34.3]}

Explanation:

Given: `f(x)=(-x+1)(x+2)(x-3)^2`
Required the zeros?
Solution Strategy: Since `f(x)` is given by a product of three function `h(x), g(x), v(x)` such `h(x)=(-x+1), g(x)=(x+2), v(x)=(x-3)^2`
`f(x)=0` for any `x` that sets either `(h(x) or g(x) or v(x))=0` thus
`h(x) = 0, x=1`
`g(x) = 0, x=-2`
`v(x) = 0, x_(1,2)=3`

There are four zero in total: at `x=-2, x=1, x=3 and x=3`