## how to solve this equation ?

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Topic review
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Every formula have to stat with $$and ends with$$.
a^{bc} is displayed as: abc
a_ is displayed as: a1

### Expand view Topic review: how to solve this equation ?

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.
Could you write the equation, please?

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