<s>[INST] 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.
 [/INST] You are working too hard. Note that
$$8x^3+27=0\iff x^3=\frac{-27}{8}\iff x=-\sqrt[3]{27/8}\iff x=-\frac{3}{2}$$
and so the only real solution is $x=-3/2$.
</s>