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


$$(2x)^3-(-3)^3$$ $$(2x-(-3))\cdot ((2x)^2+(2x(-3))+(-3)^2)$$ $$(2x+3)(4x^2-6x+9)$$
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.

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Also known as a "quadratic." Have you heard of the quadratic formula or polynomial discriminants? The given quadratic may not have real solutions mind you ... which can be determined with great speed if one knows about complex numbers and roots of unity, but perhaps that's for another day. – anon Jul 16 '12 at 21:46
The quadratic formula would tell you that your quadratic has no real roots. Note it's easier to observe that your equation is equivalent to the equation $x^3=-27/8$. – David Mitra Jul 16 '12 at 21:46
So, are you saying I should take the polynomial and get the discriminate? – Austin Broussard Jul 16 '12 at 21:47
$\sqrt{b^2-4ac}$ is what i should use for the polynomial? – Austin Broussard Jul 16 '12 at 21:47
The only real solution is what you've found. – hjpotter92 Jul 16 '12 at 21:49
up vote 12 down vote accepted

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 $x=-3/2$.

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Why is this the only real solution? – The Chaz 2.0 Jul 16 '12 at 22:06
@TheChaz Real cube roots of real numbers are unique, since $x\mapsto x^3$ is a bijection from $\mathbb R$ to $\mathbb R$. – Alex Becker Jul 16 '12 at 22:14
Sorry, I should have been a little more specific. For the sake of pedagogy, what precalculus explanation is there for why this equation has one solution, but something like $x^3-6x^2+11x-6$ has three real solutions? – The Chaz 2.0 Jul 16 '12 at 22:24
@TheChaz Well we can look at the graph of $x^3$. Or equivalently note that $x^3$ is an increasing function, so $y=x^3$ can only have one solution. I think uniqueness of cube-roots is generally taught in precalc (it was in my class, and I do not think very highly of that class). – Alex Becker Jul 16 '12 at 22:30
Thanks! Sorry to hear about your precalculus experience :) Surely this answer will help the OP have a better experience. – The Chaz 2.0 Jul 16 '12 at 22:33

Using the discriminant of $ax^3+bx^2+cx+d=8x^3+27$:

$$\Delta = 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2=0-0+0-0-27(8^2)(27^2) < 0$$

Thus there is only one real root, and that is, as Alex Becker found, $x=-3/2$.

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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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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.

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