OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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.

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

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

6 years ago
OpenStudy (anonymous):

\[x ^{3}+1^{3}= (x+1)(x2-x+1)\]

6 years ago
OpenStudy (anonymous):

\[x^{4}-x ^{3}+x-1 = (x+1)(x-1)(x ^{2}-x+1)=0\]

6 years ago
OpenStudy (anonymous):

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

6 years ago