# What are all the possible rational zeros for f(x)=32x^3-52x^2+17x+3 and how do you find all zeros?

Aug 22, 2016

$f \left(x\right)$ has zeros $1$, $- \frac{1}{8}$ and $\frac{3}{4}$

#### Explanation:

$f \left(x\right) = 32 {x}^{3} - 52 {x}^{2} + 17 x + 3$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are 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 $32$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{32} , \pm \frac{1}{16} , \pm \frac{3}{32} , \pm \frac{1}{8} , \pm \frac{3}{16} , \pm \frac{1}{4} , \pm \frac{3}{8} , \pm \frac{1}{2} , \pm \frac{3}{4} , \pm 1 , \pm \frac{3}{2} , \pm 3$

That is rather a lot of possibilities to try, but note that the sum of the coefficients of $f \left(x\right)$ is zero. That is:

$32 - 52 + 17 + 3 = 0$

So $f \left(1\right) = 0$, $x = 1$ is a zero and $\left(x - 1\right)$ is a factor:

$32 {x}^{3} - 52 {x}^{2} + 17 x + 3 = \left(x - 1\right) \left(32 {x}^{2} - 20 x - 3\right)$

We can find zeros of the remaining quadratic factor using an AC method:

Find a pair of factors of $A C = 32 \cdot 3 = 96$ which differ by $B = 20$.

The pair $24 , 4$ works. Use this pair to split the middle term, then factor by grouping:

$32 {x}^{2} - 20 x - 3$

$= 32 {x}^{2} - 24 x + 4 x - 3$

$= \left(32 {x}^{2} - 24 x\right) + \left(4 x - 3\right)$

$= 8 x \left(4 x - 3\right) + 1 \left(4 x - 3\right)$

$= \left(8 x + 1\right) \left(4 x - 3\right)$

Hence zeros: $x = - \frac{1}{8}$ and $x = \frac{3}{4}$