# What are all the possible rational zeros for y=3x^3+10x^2+4x-8 and how do you find all zeros?

Sep 7, 2016

Zeros of $y = 3 {x}^{3} + 10 {x}^{2} + 4 x - 8$ are $2$ and $\frac{2}{3}$.

#### Explanation:

If we put $x = - 2$ in $y = 3 {x}^{3} + 10 {x}^{2} + 4 x - 8$, we get

$y = 3 {\left(- 2\right)}^{3} + 10 {\left(- 2\right)}^{2} + 4 \left(- 2\right) - 8$

= 3×(-8)+10×4-8-8

= $- 24 + 40 - 8 - 8 = 0$

Hence $\left(x + 2\right)$ is a factor of $y = 3 {x}^{3} + 10 {x}^{2} + 4 x - 8$ and

$y = 3 {x}^{3} + 10 {x}^{2} + 4 x - 8$

= $3 {x}^{3} + 6 {x}^{2} + 4 {x}^{2} + 8 x - 4 x - 8$

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

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

= $\left(x + 2\right) \left(3 {x}^{2} + 6 x - 2 x - 4\right)$

= $\left(x + 2\right) \left(3 x \left(x + 2\right) - 2 \left(x + 2\right)\right)$

= $\left(x + 2\right) \left(x + 2\right) \left(3 x - 2\right)$

Hence zeros of $y = 3 {x}^{3} + 10 {x}^{2} + 4 x - 8$ are $2$ and $\frac{2}{3}$.