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I factored $x^3 - 8 = 0$ and I only got $x = 2$, but the answer said it's $x=2$ and $x=-1 \pm \sqrt{3}i$? How do you get $x=-1 \pm\sqrt{3}i$?

Michael Albanese
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Joshua Ree
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    There are complex roots.

    $x^3-8 = (x-2)(x^2+2x+4)$

    The latter has the roots you are looking for.

     – Alexander Vlasev Oct 08 '13 at 03:01
  • This is the difference of two cubes factorization (since $x^3$ and $8$ are perfect cubes), which says $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

    Using the fact that $i = \sqrt{-1}$, you can use the latter polynomial in the quadratic equation to get the complex roots.

     – MT_ Oct 08 '13 at 03:43

2 Answers2

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First of all, just a comment about the terminology. You can factor a polynomial, you can't factor an equation. What you are trying to do is solve the equation $x^3 - 8 = 0$.

As $2$ is a zero of the polynomial, $(x-2)$ is a factor by the factor theorem. So $x^3 - 8 = (x-2)p(x)$ for some polynomial. As the degree of a product of non-zero polynomials is the sum of their degrees, we see that $p(x)$ is a quadratic, so $p(x) = ax^2 + bx + c$. Comparing coefficients, we can then deduce the values of $a$, $b$, and $c$ which gives $x^3 - 2 = (x-2)(x^2+2x+4)$.

Now $x^3 - 8 = 0$ if $x - 2 = 0$ or $x^2 + 2x + 4 = 0$ by the null factor law. Solving the second equation (by using the quadratic formula for example), you find $x = -1\pm i\sqrt{3}$.

Alternatively, if you know some complex analysis (in particular, De Moivre's Theorem), you can obtain all three solutions at once.

Michael Albanese
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Let $x = y-1$.

Then $x^3 - 8 = y^3 - 3y^2 + 3y - 9 = (y-3)(y^2 + 3)$.

Setting this last expression equal to zero, we find all three roots:

One is $y = 3$ and the other two are $y = \pm i\sqrt{3}$.

Since $x = y-1$, we conclude the roots are $x = 2$ and $x = \pm i\sqrt{3} - 1$.

"QED"

Benjamin Dickman
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  • You made a mistake. All your zeroes are real. – Michael Albanese Oct 08 '13 at 04:16
  • Could you explain how you knew to choose the substitution $x = y - 1$? More importantly, how one would know to choose that substitution if the zeroes weren't already known. – Michael Albanese Oct 08 '13 at 04:19
  • I chose the substitution based on knowing the answers already, since the OP provided them. Personally, I would go from $x^3 = 8$ to $x = 2$ and find the other two answers by rotating in the complex plane (multiplying by the imaginary cube roots of unity). But mine are all silly solutions; your approach is the best one, I should think. – Benjamin Dickman Oct 08 '13 at 04:25
  • That's what I thought. Would have been nice if there was some way to figure out a nice substitution without knowing the zeroes. – Michael Albanese Oct 08 '13 at 04:27
  • @MichaelAlbanese You could imagine the roots of $x = 2$ and the other two rotated by $120$ degrees in either direction. Since the distance from the origin to the imaginary roots is $2$, draw these line segments in and note they each form $30$ degree angles with the $x$-axis. Since $\sin(30) = 1/2$, you can conclude that they are each $1$ below the $x$-axis. So shift up by $1$ to maintain one root on the $y$-axis and land the other two roots on the $x$-axis. Of course, this is all madness. – Benjamin Dickman Oct 08 '13 at 04:36