# What are all the possible rational zeros for f(x)=5x^3-9x^2+3x+1 and how do you find all zeros?

Aug 26, 2016

#### Explanation:

In synthetic division, you operate on the coefficients of each x-term (using a zero for any missing power of x).

Write the coefficients in order, like this:

5 -9 3 1
then draw an inverted long-division bracket underneath them.

It will look like this, but these are clearly not the coeff's of your function. This image is from the Purple Math website, which has an excellent description of how to do synthetic division:
http://www.purplemath.com/modules/synthdiv.htm

Check it out, especially page 2 of 4.
I can't get the drawings I made to copy here, but it's all there at Purple Math.

Aug 26, 2016

$f \left(x\right)$ has zeros $1$, $1$ and $- \frac{1}{5}$

#### Explanation:

$f \left(x\right) = 5 {x}^{3} - 9 {x}^{2} + 3 x + 1$

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 $1$ and $q$ a divisor of the coefficient $5$ of the leading term.

That means that the only possible rational zeros are:

$\pm \frac{1}{5} , \pm 1$

Note that the sum of the coefficients of $f \left(x\right)$ is zero. That is:

$5 - 9 + 3 + 1 = 0$

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

$5 {x}^{3} - 9 {x}^{2} + 3 x + 1 = \left(x - 1\right) \left(5 {x}^{2} - 4 x - 1\right)$

The sum of the coefficients of the remaining quadratic is also zero. That is:

$5 - 4 - 1 = 0$

Hence $x = 1$ is a zero of the quadratic and $\left(x - 1\right)$ a factor:

$5 {x}^{2} - 4 x - 1 = \left(x - 1\right) \left(5 x + 1\right)$

Hence the remaining zero is $x = - \frac{1}{5}$