how do you solve for x? x^4-x^3+x-1=0

Question

anonymous · Answer

by using sythetic division, you can solve it. look at the attachment. and this is the wikipedia's article about synthetic division http://en.wikipedia.org/wiki/Synthetic_division

anonymous · Answer

You can't solve through SD if you don't have anything to divide by, can you?

anonymous · Answer

take x^3 common from the first two terms and then take (x-1) common then i think u can solve it easily and the ans will be x=1,-1.

anonymous · Answer

The book answer has something completely different, it has $$\pm1, 1/2\pm \sqrt{3}/2\times i$$

anonymous · Answer

How do you get the second part? Or should I just ignore it?

anonymous · Answer

$$x ^{4}-x ^{3}+x-1=x ^{3}(x-1)+1(x-1)=(x^{3}+1)(x-1)=0$$
$$(x^{3}+1)(x-1)=0$$
$$x =1 $$
$$(x^{3}+1^{3})=0$$
$$a ^{3}+b ^{3} = (a+b)(a ^{2}-ab+b ^{2})$$

anonymous · Answer

$$x ^{3}+1^{3}= (x+1)(x2-x+1)$$

anonymous · Answer

$$x^{4}-x ^{3}+x-1 = (x+1)(x-1)(x ^{2}-x+1)=0$$

anonymous · Answer

x = 1, x=-1
use quadratic formula to find the other two roots