# 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?

Dec 3, 2016

The zeros of $f \left(x\right)$ are $\frac{2}{3}$ (with multiplicity $1$) and $- 1$ with multiplicity $2$

#### Explanation:

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

$3 {x}^{3} + 4 {x}^{2} - x - 2 = \left(3 x - 2\right) \left({x}^{2} + 2 x + 1\right)$

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

${x}^{2} + 2 x + 1 = {\left(x + 1\right)}^{2}$

So the other zeros are both $- 1$