# math

(x^3+8)/(x^4-16).

thanks.

The denominator is cleary the differerence of two squares (X^2-4)(x^2+4). That can again be factored to (x+2)(x-2)(x^2+4).

The numerator can be factored using the trinomial simplification..

http://www.algebra-online.com/sum-difference-cubes-1.htm

so (x^2-2x+4)/((x-2)(x^2+4)) is as far as I can go?

yes.

not true

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