<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] Completion of the square yields the following.
$$4x^2 - 6x + 9 = 4(x^2 - 3/2 x) + 9 = 4(x^2-3/2x + 9/16) + 27/4 = 4(x-3/4)^2 + 27/4.$$
This definitively shows your residual quadratic can have no real roots, since its graph never goes below the line $y = 27/4.$  This representation will allow you to find the complex ones easily, if you so wish.
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