Question

How to factor 2x^3-16 over 2x^2-8?

Answer 1

\[{2x^3-16}\over{2x^2-8}\]factor out the two:\[{2(x^3-8)}\over{2(x^2-4)}\]the two's cancel out:\[{\cancel{2}(x^3-8)}\over{\cancel{2}(x^2-4)}\]the numerator is a difference of cubes and the denominator is a perfect square binomial:\[{(x-2)(x^2+2x+4)}\over{(x-2)(x+2)}\]the (x-2)'s cancel out:\[{\cancel{(x-2)}(x^2+2x+4)}\over{\cancel{(x-2)}(x+2)}\]\[\large{\implies{{(x^2+2x+4)}\over{(x+2)}}}\]

Answer 2

@ashleyt2013 hope this helps!