>>12882706Sure, sorry I can't latex it.
The first polynomial only has even powers so you can substitute t=x^2 and get the easier equation
t^2-8t+4 = 0
You just solve this with the quadratic formula, delta is 48, so t=4+-2sqrt(3) if I dit it correctly
You now substitute back and get two equations
x^2 = 4+2sqrt(3)
x^2 = 4-2sqrt(3)
Now you solve to get four values of x
In this example maybe complex numbers don't even show up, hope I didn't confuse you, but the thing with 4th degree polynomials is even if you get complex roots when you factor completely, you can multiply and get an incomplete but real factorization like that in your pic
You polynomial now looks like this
(x-a)(x-b)(x-c)(x-d), where a, b, c and d are the solutions to the two equations involving x^2
You now pair up the right pairs, for example you multiply (x-a)(x-c) and so on, and you get two real second degree polynomials
The "right pairs" means when you have complex roots they're gonna be in conjugate pairs, like a+ib and a-ib, and if so you have to join this pairs
If you need help checking numbers just plug it into
wolframalpha.comHope this helps
