Find all real solutions to $8x^3+27=0$ 
Find all real solutions to $8x^3+27=0$

$(a-b)^3=a^3-b^3=(a-b)(a^2+ab+b^2)$
$$(2x)^3-(-3)^3$$ $$(2x-(-3))\cdot ((2x)^2+(2x(-3))+(-3)^2)$$ $$(2x+3)(4x^2-6x+9)$$
Now, to find solutions you must set each part $=0$. The first set of parenthesis is easy $$(2x+3)=0 ; x=-\left(\frac{3}{2}\right)$$
But, what I do not know is how to factor a trinominal (reverse of the FOIL method)
I know that $(a+b)(c+d)=(ac+ad+bc+bd)$. But coming up with the reverse does not make sense to me. If someone can only tell me how to factor a trinomial that would be great.

We know that $-3/2$ is a solution. Then we divide $x^3+27$ by $x+3/2$. Hence we have
$x^3+27=(x+3/2)(8x^2-12x+18)=0$. But, 8x^2-12x+18 don't have real solution.
