Question

How do you find all the zeros of `P(x) = (x-5)^2(x+2)^3(x+4)`?

Answer 1

Note that `x` is a zero if and only it causes at least one of the factors of `P(x)` to be zero. Hence zeros:

`5` (multiplicity `2`)

`-2` (multiplicity `3`)

`-4`

Explanation:

`P(x) = (x-5)^2(x+2)^3(x+4)`

`=(x-5)(x-5)(x+2)(x+2)(x+2)(x+4)`

Note that `P(x) = 0` if and only if at least one of its factors is zero.

So the only zeros are:

• `x=5` with multiplicity `2`
• `x=-2` with multiplicity `3`
• `x=-4` with multiplicity `1`

Essentially, finding the zeros of a polynomial (with their multiplicities) is the same as finding the linear factors of that polynomial. If `(x-a)` is a factor then `x=a` is a zero and vice versa.