How do you find the rest of the zeros given one of the zero c=2/3 and the function f(x)=3x^3+4x^2-x-2?

1 Answer
Dec 3, 2016

The zeros of f(x) are 2/3 (with multiplicity 1) and -1 with multiplicity 2

Explanation:

The given zero corresponds to a linear factor (3x-2), so f(x) will divide evenly as:

3x^3+4x^2-x-2 = (3x-2)(x^2+2x+1)

We can recognise the remaining quadratic factor as the perfect square trinomial:

x^2+2x+1 = (x+1)^2

So the other zeros are both -1