Solving equation

I've got equation x^4+4x^3-10x^2-28x-15=0
Online calculator factors it into three terms:
(x-3)(x+1)^2(x+5)=0
Ho do I do that?

It is like that:
15 can be divided by +1, -1,+3,-3,+5,-5,15 and -15.
For example we use the Horner's method with -1,because if you subtitute -1 at your equation it equals to 0.
Then,the equation is written (x+1) ($x^3+3x^2+13x-15$)=0.
Now,you should use the Horner's method at this equation: $x^3+3x^2+13x-15 =0$,using the appropriate root.Can you continue or shall I explain it further?

You would be an angel if you explaned this, cuz i don't get it

Ok, now i have got the next one, and i don't get this either.
x^4+12x^3+8x^2+84x-9=0
With the factor theorem i am stuck on the first divider

There are no linear factors. Where did you get this question from? Are you sure you have typed it in correctly?

Well, the equation in book is like this: (x^2+6x)^2+8(x^2+6x)=9

Well, the equation in book is like this: (x^2+6x)^2+8(x^2+6x)=9

Hmmm I note you are a new poster. Welcome.

Who should you listen to, the person with 5 posts or the person with 18,975 posts?

I have only made 1922 posts so if we disagree (as if that would ever happen ) must I be wrong?