# How do you find all the real and complex roots of x^4-5x^3+11x^2-25x+30=0?

Feb 2, 2016

There are 4 roots: $x = 2$, $x = 3$ and $x = \pm i \sqrt{5}$.

#### Explanation:

Let me show you two different ways how to find a factorization / the roots of your function.

(I) "Sharp eye"

With some experience and a "sharp eye", it's possible to find a factorization by "looking at the function sharply":

${x}^{4} - 5 {x}^{3} + \textcolor{b l u e}{11 {x}^{2}} - 25 x + 30$

... you see that a lot of terms are factors of $5$ but there is also a factor of $6$ in $30$. And $11 = 5 + 6$, so let's try to use that:

$= {x}^{4} - 5 {x}^{3} + \textcolor{b l u e}{5 {x}^{2} + 6 {x}^{2}} - 25 x + 30$

... rearrange the terms in a different order...

$= {x}^{4} + 5 {x}^{2} - 5 {x}^{3} - 25 x + 6 {x}^{2} + 30$

$= \underbrace{{x}^{4} + 5 {x}^{2}} \textcolor{w h i t e}{\times} \underbrace{- 5 {x}^{3} - 25 x} \textcolor{w h i t e}{x} + \underbrace{6 {x}^{2} + 30}$

... recognize the similarities between the three pairs and factor...

$= {x}^{2} \left({x}^{2} + 5\right) - 5 x \left({x}^{2} + 5\right) + 6 \left({x}^{2} + 5\right)$

$= \left({x}^{2} - 5 x + 6\right) \left({x}^{2} + 5\right)$

Now, you just need to factorize two quadratic terms to find your roots.

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The first one can be factorized e.g. with the quadratic formula or with the "completion of a circle" method. Or e.g. as follows:

${x}^{2} - 5 x + 6 = \left(x + a\right) \left(x + b\right) = {x}^{2} + \left(a + b\right) x + a \cdot b$

where $a + b = - 5$ and $a \cdot b = 6$.

This is true for $a = - 2$ and $b = - 3$, thus, the factorization is

${x}^{2} - 5 x + 6 = \left(x - 2\right) \left(x - 3\right)$

Thus, we gain two real roots $x = 2$ and $x = 3$.

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For the second one, we have

${x}^{2} + 5 = 0 \iff {x}^{2} = - 5$

As we know that ${i}^{2} = - 1$, this expression has two complex roots: $x = \pm i \sqrt{5}$.

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(II) "Slowly but surely"

It is absolutely no problem if you don't have the "sharp eye" needed to see the factorization from above.

It can be also done with some more effort in a slow but sure way.

(II a) Finding the first root

First of all, you should find a first root of the expression. Normally, you can try evaluating your term for $x = 1$, $x = - 1$, $x = 2$, $x = - 2$, ....

Here, we are successful with $x = 2$ since

${2}^{4} - 5 \cdot {2}^{3} + 11 \cdot {2}^{2} - 25 \cdot 2 + 30 = 16 - 40 + 44 - 50 + 30 = 0$

Thus, $x = 2$ is our first root.

(II b) Polynomial long division by (x -  root)

As next, you need to divide $\left({x}^{4} - 5 {x}^{3} + 11 {x}^{2} - 25 x + 30\right)$ by $\left(x - 2\right)$. This is how you can do it with polynomial long division:

$\textcolor{w h i t e}{\times} \left({x}^{4} - 5 {x}^{3} + 11 {x}^{2} - 25 x + 30\right) \div \left(x - 2\right) = {x}^{3} - 3 {x}^{2} + 5 x - 15$
$- \left({x}^{4} - 2 {x}^{3}\right)$
 color(white)(x) color(white)(xxxxxxx) /
$\textcolor{w h i t e}{\times \times} - 3 {x}^{3} + 11 {x}^{2}$
$\textcolor{w h i t e}{\times} - \left(- 3 {x}^{3} + 6 {x}^{2}\right)$
 color(white)(xxxx) color(white)(xxxxxxxxxx) /
$\textcolor{w h i t e}{\times \times \times \times \times \xi i} 5 {x}^{2} - 25 x$
$\textcolor{w h i t e}{\times \times \times \times x} - \left(5 {x}^{2} - 10 x\right)$
 color(white)(xxxxxxxxxxx) color(white)(xxxxxxxx) /
$\textcolor{w h i t e}{\times \times \times \times \times \times \times i} - 15 x + 30$
$\textcolor{w h i t e}{\times \times \times \times \times \times x} - \left(- 15 x + 30\right)$
 color(white)(xxxxxxxxxxxxxxx) color(white)(xxxxxxxxx) /
$\textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times \times \times x} 0$

Thus, your first step of factorizing is

$\left({x}^{4} - 5 {x}^{3} + 11 {x}^{2} - 25 x + 30\right) = \left(x - 2\right) \left({x}^{3} - 3 {x}^{2} + 5 x - 15\right)$

(II c) Finding the second root

With the term ${x}^{3} - 3 {x}^{2} + 5 x - 15$, you might recognize that the following factorization is possible:

${x}^{3} - 3 {x}^{2} + 5 x - 15 = {x}^{2} \left(x - 3\right) + 5 \left(x - 3\right) = \left({x}^{2} - 5\right) \left(x - 3\right)$

If you don't see that, you will need to repeat the procedure from (II a) and (II b), first of all testing $x = 1$, $x = - 1$, $x = 2$, ... until you succeed with $x = 3$ and then dividing

$\left({x}^{3} - 3 {x}^{2} + 5 x - 15\right) \div \left(x - 3\right) = {x}^{2} - 5$

(II d) Factorization

Thus, the factorization so far is

$\left({x}^{4} - 5 {x}^{3} + 11 {x}^{2} - 25 x + 30\right) = \left(x - 2\right) \left(x - 3\right) \left({x}^{2} + 5\right)$

The only thing left to do is finding the roots of ${x}^{2} + 5$.

${x}^{2} + 5 = 0$

${x}^{2} = - 5$

As ${i}^{2} = - 1$, we will find the roots

$x = \pm i \sqrt{5}$

In total, we have the factorization

$\left({x}^{4} - 5 {x}^{3} + 11 {x}^{2} - 25 x + 30\right) = \left(x - 2\right) \left(x - 3\right) \left(x + i \sqrt{5}\right) \left(x - i \sqrt{5}\right)$

and the roots

$x = 2$, $x = 3$ and $x = \pm i \sqrt{5}$