Question

Solve for x: [4x(x^2-4)(x^2+4)-2x(x^2+4)^2]/(x^2+4)^4

Answer 1

\[\large \frac{4x(x^2-4)(x^2+4)-2x(x^2+4)^2}{(x^2+4)^4}\]factor out an (x^2+4) and a 2x from numerator \[\large \frac{2x(x^2+4)[2x(x^2-4)-(x^2+4)]}{(x^2+4)^4}\] cancel out like terms \[\large \frac{2x \cancel{(x^2+4)}[2x(x^2-4)-(x^2+4)]}{(x^2+4)^\cancel{4}}\] the bottom exponent becomes 3. \[\large \frac{2x[2x(x^2-4)-(x^2+4)]}{(x^2+4)^3}\] simplify stuff inside brackets. \[\large \frac{2x(x^2-4-x^2-4)}{(x^2+4)^3}\]

Answer 2

the last step is written like this, i made a mistake \[\large \frac{2x[2x(x^2-4-x^2-4)]}{(x^2+4)^3}\]

Answer 3

\[\large \frac{2x[2x(-4-4)]}{(x^2+4)^3}\] simplify \[\large \frac{2x[2x(-8)]}{(x^2+4)^3}\] simplify more... \[\large \frac{-32x^2}{(x^2+4)^3}\]

Answer 4

i see... it would be \[\large \frac{2x[2x(x^2-4)-(x^2+4)]}{(x^2+4)^3}\]\[\large \frac{2x(2x^3-8x-x^2-4)}{(x^2+4)^3}\]