Doing a lot of homework tonight, I will probably be asking multiple questions throughout the next few hours

I have to find all of the factors of this polynomial -
x^4-3x^2-4
(x-2)(x+2) are factors have to find the rest

Please show work with your answer so I can understand what I'm doing wrong, I'm using synthetic division but I get a remainder

This problem only has two roots, x=2, and x=-2.

the other factor is (x^2+1) but I didn't calculate that myself, I don't know how to do it. Can anyone elaborate?

you have (x+2)(x-2)=x^2-4 now you try (x^2+Z)(x^2-4)=X^4+Z*x^2-4*x^2-Z*4 since you need Z*4=4 try with Z=1 and hurray it works
that is the quick and dirty way
now x^2+1=(x+i)(x-i) so:

X^4-3*x^2-4 = (x+2)(x-2)(x+i)(x-i)

& @



and no, it has 4 roots like every polynomial of degree 4
while sometimes you have one or more roots with multiplicity
here you have 4 distinct roots: +2,-2,+i,-i

This post was edited by brmv on Apr 6 2014 09:10pm

Sorry, I'm pretty fucking thick. (x^2+Z)(x^2-4)=X^4+Z*x^2-4*x^2-Z*4 I'm not understanding where this came from

you want to find factors of x^4 - 3*x^2 - 4 and you know that (x+2)*(x-2) = x^2 -4 is a factor

to get (x^4 + something) from (x^2-4) you will have to find a factor with x^2, ie to achieve x^4

Another way to look at this :

After you did your synthetic division and found that 2 and -2 were roots, you were left with x^(2)+1.
From here you can set this equal to zero and solve for x:

x^(2)+1 = 0
x^(2) = -1
x = +sqrt(-1) or x = -sqrt(-1) [or you could say x equals positive OR negative square root of -1]
so x = +i or x = -i

now you have all of the roots.
2 rational roots, 2 irrational roots.

BAH, I thought I typed real. I meant to say it only has two real roots.

I think you messed up: