Question 292165
Well, you have a couple of choices:
1) You could divide using long division or synthetic division.
2) You could try factoring the numerator and see if anything will cancel the denominator.
Let's try 2)
{{{(24x^3+3)/(2x+1)}}} Factor 3 from the numerator.
{{{3(8x^3+1)/(2x+1)}}} Now factor the binomial in the numerator as the sum of two cubes.
{{{3(8x^3+1) = 3(highlight_green((2x)^3+(1)^3))}}}
Recall that the sum of two cubes will factor thus:
{{{A^3+B^3 = (A+B)(A^2-AB+B^2)}}} and here, A = 2x, B = 1, so...
{{{3(8x^3+1) = 3(2x+1)(4x^2-2x+1)}}} so now we have:
{{{3cross((2x+1))(4x^2-2x+1)/cross((2x+1))}}}={{{3(4x^2-2x+1) = highlight(12x^2-6x+3)}}}