SOLUTION: a) Find x: x^5+ x^4+ x^3+ x^2+1=0 I have factored it : (x^2+ x+1)(x+1)(x^2-x+1) b)Factoring: 2x^2y^2+2y^2z^2+2x^2z^2-x^4-y^4-z^4

Algebra ->  Rational-functions -> SOLUTION: a) Find x: x^5+ x^4+ x^3+ x^2+1=0 I have factored it : (x^2+ x+1)(x+1)(x^2-x+1) b)Factoring: 2x^2y^2+2y^2z^2+2x^2z^2-x^4-y^4-z^4      Log On


   

Question 1119660: a) Find x:
x^5+ x^4+ x^3+ x^2+1=0
I have factored it : (x^2+ x+1)(x+1)(x^2-x+1)
b)Factoring:
2x^2y^2+2y^2z^2+2x^2z^2-x^4-y^4-z^4
Found 4 solutions by Alan3354, ikleyn, greenestamps, solver91311:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
a) Find x:
x^5+ x^4+ x^3+ x^2+1=0
I have factored it : (x^2+ x+1)(x+1)(x^2-x+1)
--------------
Then (x^2+ x+1)(x+1)(x^2-x+1) = 0
--> x = -1 is the most obvious solution
-----
But
f(x) = x^5+ x^4+ x^3+ x^2+1
f(-1) = -1 + 1 - 1 + 1 + 1 = 1, not zero, so (x+1) is not a factor.
======================
b)Factoring:
2x^2y^2+2y^2z^2+2x^2z^2-x^4-y^4-z^4

Answer by ikleyn(52835) About Me  (Show Source):
You can put this solution on YOUR website!
.
a)  Regarding the equation     //   !!  as it is written in your post  !!

    x^5+ x^4+ x^3+ x^2+1=0,


    your factoring is INCORRECT, since it implies that x= -1 is the root,

    while easy inspection shows it is not.


    The same arguments disprove the solution by the other tutor.



b)  2x%5E2y%5E2%2B2y%5E2z%5E2%2B2x%5E2z%5E2-x%5E4-y%5E4-z%5E4
~~~~~~~~~~~~~~~~~~~~~~


    Regarding this factoring, see this video-lecture 


    https://www.youtube.com/watch?v=SWMPwVlF07g


Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


You didn't write the first equation correctly; you left out the linear term.

x%5E5%2Bx%5E4%2Bx%5E3%2Bx%5E2%2Bx%2B1+=+0
%28x%2B1%29%28x%5E4%2Bx%5E2%2B1%29+=+0
%28x%2B1%29%28%28x%5E4%2B2x%5E2%2B1%29-x%5E2%29+=+0
%28x%2B1%29%28%28%28x%5E2%2B1%29%2Bx%29%28%28x%5E2%2B1%29-x%29%29+=+0
%28x%2B1%29%28x%5E2%2Bx%2B1%29%28x%5E2-x%2B1%29+=+0

The linear factor gives the only real solution, x = -1. Each of the quadratic factors gives a pair of complex solutions.

To factor the second expression, we can use a method similar to that used above to factor x^4+x^2+1.

2x%5E2y%5E2%2B2y%5E2z%5E2%2B2x%5E2z%5E2-x%5E4-y%5E4-z%5E4
= %28-1%29%28x%5E4%2By%5E4%2Bz%5E4-2x%5E2y%5E2-2x%5E2z%5E2-2y%5E2z%5E2%29
= %28-1%29%28%28x%5E4%2By%5E4%2Bz%5E4%2B2x%5E2y%5E2-2x%5E2z%5E2-2y%5E2z%5E2%29-4x%5E2y%5E2%29
= %28-1%29%28%28x%5E2%2By%5E2-z%5E2%29%5E2-%282xy%29%5E2%29
= %28-1%29%28x%5E2%2By%5E2-z%5E2%2B2xy%29%28x%5E2%2By%5E2-z%5E2-2xy%29

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


You have done all of the work by factoring the quintic equation. Just set each of the factors equal to zero and solve. You have a linear binomial so one of the roots is x = -1. Solve the other two with the quadratic formula. Note that four of your roots are complex.




John

My calculator said it, I believe it, that settles it