how to solve this equation ?

by **abdlaziz** » Sun Feb 14, 2010 9:36 pm

this is first time to me in this amazing forum

can any body tell me how to solve this equation?

factor this equation :

x[sup]8[/sup]+x[sup]4[/sup]+1

by **Math Tutor** » Mon Feb 15, 2010 6:49 am

Could you write the equation, please?

$x^8 + 2x^4 + 1 = x^8 + 2x^4 + 1 - x^4 =$
$= (x^4 + 1)^2 - x^4 = (x^4 + 1 - x^2)(x^4 + 1 + x^2)=$
$= (x^4 + 2x + 1 - 3x^2)(x^4 + 2x^2+1 - x^2)=$
$= ((x^2 + 1)^2 - 3x^2)((x^2 +1)^2 - x^2)=$
$= (x^2 + 1 - sqrt{3}x)(x^2 + 1 - sqrt{3}x)(x^2 +1 - x)(x^2 +1 + x)=$

May you continue, please?

by **martosss** » Sun Sep 19, 2010 10:05 am

teacher wrote: $= (x^4 + 1)^2 - x^4 = (x^4 + 1 - x^2)(x^4 + 1 + x^2)=$
$= (x^4 + 2x + 1 - 3x^2)(x^4 + 2x^2+1 - x^2)=$

Your result is correct, but you have technical mistakes!
$(x^4+1-x^2)(x^4+1+x^2)=$
$\left[(x^2-1)^2+x^2\right]\left[(x^2+\frac{1}{2})^2+\frac{3}{4}\right]$
You can also present it as
$(x^4+\frac{1}{2})^2+\frac{3}{4}$ , which has no solutions if it is equal to 0.

how to solve this equation ?

how to solve this equation ?

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