# Can I factor it?

#### Beez

On my process of obtaining the IF for solving a differential equation, I got stuck with an equation as following. This could be a very simple algebra problem, but I just can't do it. Would someone tell me if I can factor the numerators by (x^4 + 1) for the first equation and by (x^4 -1) for the second equation so that I can have only x or y as a variable?

[(x^4 - 1) - (5*x^4 + 1)]/x*(x^4+1) = (-4*x^4 - 2)/x(x^4 + 1)
= -2(2*x^4 +1)/x(x^4 + 1)

[(5*x^4 + 1) - (x^4 - 1)]/y*(x^4-1) = (4x^4 + 2)/y(x^4 - 1)
=2(2*x^4 + 1)/y(x^4 - 1)

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could you use $$\LaTeX$$ please. it is kinda hard to read without it.

#### Beez

OK, I got it. Please take look at the following.

[($$x^4$$ - 1) - (5$$x^4$$+ 1)]/x($$x^4$$+1) = (-4$$x^4$$ - 2)/x($$x^4$$ + 1)
= -2(2$$x^4$$ +1)/x($$x^4$$+ 1)

[(5$$x^4$$ + 1) - ($$x^4$$ - 1)]/y($$x^4$$-1) = (4$$x^4$$ + 2)/y($$x^4$$- 1)
=2(2$$x^4$$+ 1)/y($$x^4$$ - 1)

Thanks for your suggestion. I always wanted to know how to type $$x^2$$ for x^2.

Last edited:

ur kidding, right? one exponent isn't enough(especially without the x )

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