Question

How do you find all the zeros of `f(x)= x^4 + 3x^3 - 4x^2 - 12x` with its multiplicities?

Answer 1

Zeros: `0, 2, -2, -3`

Explanation:

`f(x) = x^4+3x^3-4x^2-12x`

Note that all of the terms are divisible by `x` and the ratio of the first and second terms is the same as that between the third and fourth terms. So this will factor by grouping...

`x^4+3x^3-4x^2-12x`

`=x((x^3+3x^2)-(4x+12))`

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

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

`=x(x^2-2^2)(x+3)`

`=x(x-2)(x+2)(x+3)`

Hence zeros:

`0, 2, -2, -3`