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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.

  • 0
    $a^3 x^3 + b^3 = 0 \implies x = -\sqrt[3]{\dfrac{b^3}{a^3}} = -\dfrac{b}{a}.$2012-07-17

4 Answers 4

12

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 $x=-3/2$.

  • 0
    Thanks! Sorry to hear about your precalculus experience :) Surely this answer will help the OP have a better experience.2012-07-16
3

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, $x=-3/2$.

2

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.

1

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.