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

##### 1 Answer

There are 4 roots:

#### 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 - 5x^3 + color(blue)(11x^2) - 25x + 30#

... you see that a lot of terms are factors of

#= x^4 - 5x^3 + color(blue)(5 x^2 + 6x^2) - 25x + 30#

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

#= x^4 + 5x^2 - 5x^3 - 25x + 6x^2 + 30#

#= underbrace(x^4 + 5x^2) color(white)(xx) underbrace(-5x^3 - 25x) color(white)(x) + underbrace(6x^2 + 30)#

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

# = x^2 (x^2 + 5) - 5x (x^2 + 5) + 6(x^2 + 5)#

# = (x^2 - 5x + 6)(x^2 + 5)#

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 - 5x + 6 = (x + a)(x + b) = x^2 + (a+b)x + a*b#

where

This is true for

#x^2 - 5x + 6 = (x-2)(x-3)#

Thus, we gain two real roots

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For the second one, we have

#x^2 + 5 = 0 <=> x^2 = -5#

As we know that

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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

Here, we are successful with

#2^4 - 5* 2^3 + 11 * 2^2 - 25 * 2 + 30 = 16 - 40 + 44 - 50 + 30 = 0#

Thus,

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

As next, you need to divide

# color(white)(xx) (x^4 - 5x^3 + 11 x^2 - 25x + 30) -: (x-2) = x^3 - 3x^2 + 5x - 15#

# -(x^4 - 2x^3)#

# color(white)(x) color(white)(xxxxxxx) / #

# color(white)(xxxx) -3x^3 + 11x^2#

# color(white)(xx) -(-3x^3 + 6x^2)#

# color(white)(xxxx) color(white)(xxxxxxxxxx) / #

# color(white)(xxxxxxxxxxxii) 5x^2 - 25x#

# color(white)(xxxxxxxxx) -(5x^2 - 10x)#

# color(white)(xxxxxxxxxxx) color(white)(xxxxxxxx) / #

# color(white)(xxxxxxxxxxxxxxi) -15x + 30#

# color(white)(xxxxxxxxxxxxx) -(-15x + 30)#

# color(white)(xxxxxxxxxxxxxxx) color(white)(xxxxxxxxx) / #

# color(white)(xxxxxxxxxxxxxxxxxxxxxxx) 0#

Thus, your first step of factorizing is

#(x^4 - 5x^3 + 11 x^2 - 25x + 30) = (x-2)(x^3 - 3x^2 + 5x - 15)#

**(II c) Finding the second root**

With the term

#x^3 - 3x^2 + 5x - 15 = x^2(x-3) + 5(x-3) = (x^2 - 5)(x-3)#

If you don't see that, you will need to repeat the procedure from (II a) and (II b), first of all testing

#(x^3 - 3x^2 + 5x - 15) -: (x - 3) = x^2 - 5#

**(II d) Factorization**

Thus, the factorization so far is

#(x^4 - 5x^3 + 11 x^2 - 25x + 30) = (x-2)(x - 3)(x^2 + 5)#

The only thing left to do is finding the roots of

# x^2 + 5 = 0#

# x^2 = -5#

As

#x = +- i sqrt(5)#

In total, we have the factorization

#(x^4 - 5x^3 + 11 x^2 - 25x + 30) = (x-2)(x - 3)(x + i sqrt(5))(x - i sqrt(5))#

and the roots

#x = 2# ,#x = 3# and#x = +- i sqrt(5)#