Question

5x^4 − 5x^3 + 5x − 5 = 0 all answers including complex

Answer 1

\[5x^{4}-5x^{3}+5x-5=0\]divide by 5:\[x^{4}-x^{3}+x-1=0\]\[x^{3}(x-1)+x-1=0\]take the common factor (x-1):\[(x-1)(x^{3}+1)=0\]so\[x-1=0\]or\[x^{3}+1=0\]giving x=1 or -1

Answer 2

{x -> -1}, {x -> 1}, {x -> (-1)^(1/3)}, {x -> -(-1)^(2/3)}

Answer 3

5x^4-5x^3+5x-5 = 0
5x^3(x-1)+5(x-1) =0
(x-1)(5x^3+5) = 0
x = 1, -1

Answer 4

\[x^3+1=0\]

then \[x=-1,\frac{1}{2}\pm\frac{\sqrt{3}}{2}\]

Answer 5

forgot my i

Answer 6

\[x=-1,\frac{1}{2}\pm\frac{\sqrt{3}}{2}i\]

Answer 7

Of course - I completely forgot about the complex solutions - thanks Zarkon!

Answer 8

What? I'm slow. cube root of -1 = -1? How do you get complex numbers from that?

Answer 9

Zarkon, explain

Answer 10

\[x^3+1\] is a cubic polynomial and therefore is must have 3 roots (counting multiplicity)


you can find the roots by looking at the unit disk.


I'll draw a picture

Answer 11

what a crappy picture ;)

the roots lie on the unit disk(in the complex plane) 360/3=120 degrees apart

Answer 12

you can find them by using

\[\cos(\pi/3)+\sin(\pi/3)i\]
and
\[-(\cos(2\pi/3)+\sin(2\pi/3)i)\]

Answer 13

Okay, I forgot about that, thanks.

Answer 14

\[(1/2+i\sqrt{3}/2)^{2}=-1/2+i\sqrt{3}/2\]so\[(1/2+i\sqrt{3}/2)^{3}=(1/2+i\sqrt{3}/2)^{2}*(1/2+i\sqrt{3}/2)=(-1/2+i\sqrt{3}/2)*(1/2+i\sqrt{3}/2)^{2}\]\[=-1/4-3/4=-1\]