Factor completely:x^4+4
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March 16th, 2008
I got two answers, (x^2+2x+2)(x^2–2x+2) and (x^2+2i)(x^2–2i). I think both work. Are they both correct, and how do you get the second one?
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7 Answers
yeah, they both work, I checked it. Unless I made some stupid mistake… I’m not sure how you get the imaginary one though I think it has something to do with factoring your first answer further and distributing it some funky way…. I’m trying to work it out now. :)
Yes, the four solutions to x^4+4=0 are x=±1±i. So to factor this completely, you have:
x^4+4=(x+1+i)(x+1-i)(x-1+i)(x-1-i)
If you multiply the first two factors and the second two factors, you obtain
x^4+4=(x^2+2x+2)(x^2–2x+2)
Alternatively, if you multiply the first and last factor, and also the second and third, you get:
x^4+4=(x^2–2i)(x^2+2i)
You can, of course, write the factorization in a third way, that being
x^4+4=(x^2+2ix-2)(x^2–2ix-2).
Thanks for the responses!
How do you factor it completely? [How do you get (x+1+i)(x+1-i)(x-1+i)(x-1-i)]
you get that by factoring (x^2+2x+2) and (x^2–2x+2) using the quadratic formula for each.
For the first one:
(- 2 +- (2^2 – 4(1)(2))^1/2)/2
= (- 2 +-(4–8)^1/2))/2
= (- 2 +- (-4)^1/2)/2
=(- 2 +-2i)/2
=-1 +-i
and follow the same procedure for the other one. :)
@chris, how do u do the +- thing? lol
@ 8lightminutesaway
If you have a mac, you get ± by pressing the (+/=) key while holding (option + shift). I’m afraid I can’t help for a pc.
A quicker way to do this equation in particular is to notice that you’re looking for the fourth roots of -4. When you find one, the others have to be that times the four fourth roots of one (namely, 1, -1, i, and -i). So once you find one, you’ve found them all.
There are quick ways to geometrically see where the fourth roots of -4 are if you look at the complex plane and use polar coordinates, but it’s a little tricky to describe without being able to draw pictures.
Those are both correct answers. Good job!
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