# What are all the possible rational zeros for f(x)=x^3-4x^2-14 and how do you find all zeros?

Jan 24, 2017

We expect the rational zeros (i.e. the roots) to be factors of the constant, meaning that the possible rational roots are:

$\pm 1 , \pm 2 , \pm 7 , \pm 14$

We would literally test all the roots, until we find three that work. There are only as many roots as the highest degree in the polynomial.

Let's try $x = \pm 1 , \pm 2 , \pm 7$ first. Set up a synthetic division by examining the coefficients in your cubic. Assume the coefficient for $x$ is there, but $0$.

ul(1" ") "|"" "" "1" "" "-4" "" "0" "" "" "-14
" "+ " "ul(" "" "" "" "" "" "" "" "" "" "" "" "" ")

The general steps are:

1. Bring down the first coefficient.
2. Multiply the result beneath the horizontal line by the root and store in the next column above the horizontal line.
4. Repeat 2-3 until you reach the last column and have evaluated the final addition.

And you should get:

ul(1" ") "|"" "" "1" "" "-4" "" "0" "" "" "-14
" "+ " "ul(" "" "" "" "1" "-3" "" "" "-3" "" ")
$\text{ "" "" "" "1" "-3" "-3" "" "" } \textcolor{red}{- 17}$

This didn't work, because the final column ends with a nonzero value. That means the remainder is $- 17$ and the division leaves you with a $- \frac{17}{x - 1}$. So, there was not a clean division. Now is the time to try other roots!

Below are other tries. Highlighted in red is the remainder if it is not zero.

ul(2" ") "|"" "" "1" "" "-4" "" "0" "" "" "-14
" "+ " "ul(" "" "" "" "2" "-4" "" "" "-8" "" ")
$\text{ "" "" "" "1" "-2" "-4" "" "" } \textcolor{red}{- 22}$

ul(7" ") "|"" "" "1" "" "-4" "" "0" "" "" "-14
" "+ " "ul(" "" "" "" "7" "" "" "21" "" "" "147" "" ")
$\text{ "" "" "" "1" "" "3" "" "" "21" "" "" } \textcolor{red}{133}$

ul(-1" ") "|"" "" "1" "" "-4" "" "0" "" "" "-14
" "+ " "ul(" "" "" "" "-1" "" "5" "" "" "" "-5" "" ")
$\text{ "" "" "" "1" "" "-5" "" "5" "" "" } \textcolor{red}{- 19}$

ul(-2" ") "|"" "" "1" "" "-4" "" "0" "" "" "-14
" "+ " "ul(" "" "" "" "-2" "" "12" "" "-24" "" ")
$\text{ "" "" "" "1" "" "-6" "" "12" "" "" } \textcolor{red}{- 38}$

ul(-7" ") "|"" "" "1" "" "-4" "" "0" "" "" "-14
" "+ " "ul(" "" "" "" "-7" "" "77" "" "" "-539" "" ")
$\text{ "" "" "" "1" "" "-11" "" "77" "" "" } \textcolor{red}{553}$

I hate to say it dude, but there are no rational zeros, so you'd be wasting your time trying to find any. It turns out that by numerical methods, you'd get the three roots as:

$\textcolor{b l u e}{{r}_{1} = \frac{4}{3} + \frac{1}{3} {\left(253 - 3 \sqrt{6657}\right)}^{\text{1/3") + 1/3(253 + 3 sqrt(6657))^("1/3}} \approx 4.6480}$

$\textcolor{b l u e}{{r}_{2} , {r}_{3} = - 0.3240 \pm 1.7050 i}$