# Help! Find all zeros: f(x)=3x^7-32x^6+28x^5+591x^4-1181x^3-2810x^2+5550x-1125?

Oct 16, 2017

$x = - 3 , - 3 , 5 , 5 , 5 , \frac{5 + \sqrt{13}}{6} , \frac{5 - \sqrt{13}}{6}$

#### Explanation:

This one is extremely...tedious...to solve. You really need some luck and clever ideas on how to proceed.

First, I apply the Rational Roots Theorem to come up with a list of all possible rational roots for $f \left(x\right) = 0$. I list all of the factors of the constant term $p$ and the leading coefficient term $q$, and then consider all reduced form rational terms of the form $\pm \left(p \text{ factor")/(q " factor}\right)$:

$p \text{ factors} : 1 , 3 , 5 , 9 , 15 , 25 , 45 , 75 , 125 , 225 , 375 , 1125$

q " factors": 1, 3

Possible rational roots:

$\pm \left(1 , \frac{1}{3} , 3 , 5 , \frac{5}{3} , 9 , 15 , 25 , \frac{25}{3} , 45 , 75 , 125 , \frac{125}{3} , 225 , 375 , 1125\right)$

This is fairly huge. There are 32 possible rational roots. Since this is a degree 7 polynomial, there will be 7 roots (although some may be complex or irrational in nature).

From here, it's trial and error time. Synthetic division can help; test a possible root using synthetic division and look for a final value of 0, which would indicate you've located a zero. I'll omit all the trials that did not succeed for me to save space.

Try x = -3

$- 3 \rfloor \textcolor{w h i t e}{\text{a")3color(white)("aa")-32color(white)("aa")28color(white)("aaa")591color(white)("aa")-1181color(white)("aa")-2810color(white)("aaa")5550color(white)("a}} - 1125$
$\underline{\textcolor{w h i t e}{\text{aaaaaaaaa")-9color(white)("a")123color(white)("")-453color(white)("aaa")-414color(white)("aaaaa")4785color(white)("")-5925color(white)("aaaa}} 1125}$
$\textcolor{w h i t e}{\text{aaaa")3color(white)("aa")-41color(white)("a")151color(white)("aaa")138color(white)("aa")-1595color(white)("aaaa")1975color(white)("aa")-375color(white)("aaaaa}} \textcolor{red}{0}$

This result tells us that $x = - 3$ is a zero. From now on, we can use the reduced final row for our synthetic testing.

Try x= 5

$5 \rfloor \textcolor{w h i t e}{\text{aaaa")3color(white)("aa")-41color(white)("aaa")151color(white)("aaa")138color(white)("aa")-1595color(white)("aaaa")1975color(white)("aa}} - 375$
underline(color(white)("aaaaaaaaaaa")15color(white) ("a")-130color(white)("aaa")105color(white)("aaaa")1215color(white)("aa")-1900color(white)("aaaa")375)
$\textcolor{w h i t e}{\text{aaaaaa")3color(white)("aa")-26color(white)("aaaa")21color(white)("aaa")243color(white)("aa")-380color(white)("aaaaaaa")75color(white)("aaaaa}} \textcolor{red}{0}$

This result tell us that $x = 5$ is a zero. Ignoring the final 0 result, we are now down to a 5th degree polynomial.

The next step is a trick. We already know $x = - 3$ is a zero. It could be a zero of multiplicity; in other words, it could occur as more than one factor.

Try x = -3 (again)

$- 3 \rfloor \textcolor{w h i t e}{\text{aaaa")3color(white)("aa")-26color(white)("aaaa")21color(white)("aaaa")243color(white)("aa")-380color(white)("aaaaaa}} 75$
underline(color(white)("aaaaaaaaaaaa")-9color(white) ("aaaa")105color(white)("a")-378color(white)("aaaa")405color(white)("aaaa")-75)#
$\textcolor{w h i t e}{\text{aaaaaaaa")3color(white)("aa")-35color(white)("aaa")126color(white)("a")-135color(white)("aaaaa")25color(white)("aaaaaaa}} \textcolor{red}{0}$

This result tells us that $x = - 3$ is a zero a 2nd time. An additional try of -3 (not shown here) shows that -3 is not a zero a 3rd time. (An application of Descarte's Rule of Signs would demonstrate that -3 could not be a zero three times.)

Try x = 5 (again)

$5 \rfloor \textcolor{w h i t e}{\text{aaaaa")3color(white)("aaa")-35color(white)("aaaa")126color(white)("aaa")-135color(white)("aaaaa}} 25$
$\underline{\textcolor{w h i t e}{\text{aaaaaaaaaaaaa")15color(white)("aa")-100color(white)("aaaaa")130color(white)("aaa}} - 25}$
$\textcolor{w h i t e}{\text{aaaaaaa")3color(white)("aaa")-20color(white)("aaaaa")26color(white)("aaaa")-5color(white)("aaaaaa}} \textcolor{red}{0}$

This result tells us that $x = 5$ is a zero a 2nd time. Could it be a zero a third time?

Try x = 5 (again x 2)

$5 \rfloor \textcolor{w h i t e}{\text{aaaa")3color(white)("aaa")-20color(white)("aaaaa")26color(white)("aaaa}} - 5$
$\underline{\textcolor{w h i t e}{\text{aaaaaaaaaaaa")15color(white)("aaa")-25color(white)("aaaaaa}} 5}$
$\textcolor{w h i t e}{\text{aaaaaa")3color(white)("aaa")-5color(white)("aaaaaa")1color(white)("aaaaaaa}} \textcolor{red}{0}$

This result tells us that $x = 5$ is a zero a 3rd time.

At this point we are now left with a quadratic polynomial of $3 {x}^{2} - 5 x + 1$. which we can complete using the Quadratic Formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{5 \pm \sqrt{{\left(= 5\right)}^{2} - 4 \left(3\right) \left(1\right)}}{2 \cdot 3}$

$x = \frac{5 \pm \sqrt{25 - 12}}{6} = \frac{5 \pm \sqrt{13}}{6}$

Our final collection of all 7 roots is:

$x = - 3 , - 3 , 5 , 5 , 5 , \frac{5 + \sqrt{13}}{6} , \frac{5 - \sqrt{13}}{6}$