# What are all the possible rational zeros for f(x)=3x^3+11x^2+5x-3 and how do you find all zeros?

Sep 7, 2016

The Zeroes of $f$ are 1/3,-3,&,-1.

#### Explanation:

We will factorise $f \left(x\right) = 3 {x}^{3} + 11 {x}^{2} + 5 x - 3 ,$ to find its zeroes.

Observe that,

The Sum of the co-effs. of Odd-powered terms$= 3 + 5 = 8 ,$ and,

that of the Even-powered ones$= 11 - 3 = 8$.

Hence, $\left(x + 1\right)$ is a factor of $f \left(x\right)$.

Now, $f \left(x\right) = 3 {x}^{3} + 11 {x}^{2} + 5 x - 3$

$= \underline{3 {x}^{3} + 3 {x}^{2}} + \underline{8 {x}^{2} + 8 x} - \underline{3 x - 3}$

$= 3 {x}^{2} \left(x + 1\right) + 8 x \left(x + 1\right) - 3 \left(x + 1\right)$

$= \left(x + 1\right) \left(3 {x}^{2} + 8 x - 3\right)$

$= \left(x + 1\right) \left\{\underline{3 {x}^{2} + 9 x} - \underline{x - 3}\right\}$

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

$= \left(x + 1\right) \left(x + 3\right) \left(3 x - 1\right)$

Hence, the Zeroes of $f$ are 1/3,-3,&,-1.