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

Oct 26, 2016

$f \left(x\right)$ has zeros $1$, $- \frac{1}{3}$ and $3$

#### Explanation:

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

By the rational roots theorem, any rational zero of $f \left(x\right)$ is expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $3$ and $q$ a divisor of the coefficient $3$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{3} , \pm 1 , \pm 3$

Note first that there's a shortcut in this particular example, in that the sum of the coefficients is $0$. That is $3 - 11 + 5 + 3 = 0$

Hence we can deduce that $f \left(1\right) = 0$, $x = 1$ is a zero of $f \left(x\right)$ and $\left(x - 1\right)$ is a factor:

$3 {x}^{3} - 11 {x}^{2} + 5 x + 3 = \left(x - 1\right) \left(3 {x}^{2} - 8 x - 3\right)$

We could just try the remaining rational possibilities that we listed, but we can factor the remaining quadratic ($3 {x}^{2} - 8 x - 3$) easily enough using an AC method.

Find a pair of factors of $A C = 3 \cdot 3 = 9$ which differ by $B = 8$.

The pair $9 , 1$ works in that $9 \cdot 1 = 9$ and $9 - 1 = 8$

Use this pair to split the middle term and factor by grouping:

$3 {x}^{2} - 8 x - 3 = 3 {x}^{2} - 9 x + x - 3$

$\textcolor{w h i t e}{3 {x}^{2} - 8 x - 3} = \left(3 {x}^{2} - 9 x\right) + \left(x - 3\right)$

$\textcolor{w h i t e}{3 {x}^{2} - 8 x - 3} = 3 x \left(x - 3\right) + 1 \left(x - 3\right)$

$\textcolor{w h i t e}{3 {x}^{2} - 8 x - 3} = \left(3 x + 1\right) \left(x - 3\right)$

Hence the other two zeros of $f \left(x\right)$ are $x = - \frac{1}{3}$ and $x = 3$