Factorising complex equations

Question

I do factorization by just plugging in simple numbers to check whether they are factors, using long division and also using synthetic division. But, how to do it in the case of complex equations. For eg: Factorize $x^6+5x^3+8$

Putting $v=x^3$ I get $v^2+5v+8$. But, this equation has complex roots. So, how to continue?

Answer 1

To solve $v^2+5v+8=0$, observe that the equation is equivalent to $\left(v+\frac{5}{2}\right)^2=-8+\frac{25}{4}=-\frac{7}{4}$. Solve this for $v+\frac{5}{2}$.

Answer 2

I'm sorry, I didn't read the question properly. When you say factorize, do you mean into polynomials with rational coefficients? If the coefficients may be irrational, then in principle you can use what I wrote above to find all 6 complex zeros of the polynomial. Because the polynomial has real coefficients, the zeros pair up in conjugates. Indeed $(x-(a+bi))(x-(a-bi))=x^2-2xa+a^2+b^2$ is real ($a,b \in \mathbb{R}$), and so you may write the polynomial as a product of 3 such conjugate factors (http://www.wolframalpha.com/input/?i=factorize+x%5E6%2B5x%5E3%2B8, http://www.wolframalpha.com/input/?i=solve+x%5E6%2B5x%5E3%2B8%3D0)

If the coefficients must be rational,you will have to look at what $a$ and $b$ in a solution $a+bi$ will make $2a$ and $a^2+b^2$ rational. In this case, inspection shows (referring to the second link above) that the only possibility is $a=1/2$, $b=\pm \sqrt{7}/2$. This gives a factor $x^2-x+2$, and using polynomial division, you find $x^6+5x^3+8=(x^2-x)(x^4+x^3-x^2+2x+4)$.

Answer 3

Formally, you can write $a,\overline a$ for your pair of complex roots of the equation for $v$ and take complex cube roots $\alpha0,\alpha_1\alpha_2$ of $a$, in which case you get $$ x^6+5x^3+8 = (x-\alpha_0)(x-\alpha_1)(x-\alpha_2)(x-\overline{\alpha_0})(x-\overline{\alpha_1})(x-\overline{\alpha_2}) $$ as concrete factorisation of your polynomial. Concretely one has $a=\frac{-5+\mathbf i\sqrt7}2$, and $\alpha_i=\alpha_0\exp(\frac{2k}3\pi\mathbf i)\alpha_0$ for $k=1,2$, but this still does not make the (chosen) cube root $\alpha_0$ of$~a$ explicit.

To do "better", you could start to try to factor your polynomial over the rational numbers first. Since the above at least shows that there are no real roots, the best to hope for is a factorisation involving an irreducible quadratic factor (if there were a cubic factor, there would have been real roots). Since the given polynomial has integer coefficients with $\gcd$ equal to $1$ (since it is monic), any rational factors can be taken to have this property as well (due to Gauss's lemma). So we are looking for a factor $x^2+bx+c$ with $b^2<4c$ (to exclude real roots) and $c\in\{-8,-4,-2,-1,1,2,4,8\}$ (as it must divide$~8$). In order that the coefficients of $x^6,x^5,x^4,x^1,x^0$ in the product become what they should be, the other factor can only be $x^4-bx^3+(b^2-c)x^2-\frac{8b}{c^2}x+\frac8c$. Computing the coefficient of $x^2$ in the product and multiplying by $c^2$ we find $b^2c^3-c^4-8b^2+8c=(b^2-c)(c^3-8)$, which must be zero, so either $(b,c)\in\{(1,1),(2,4)\}$ or $c=2$. In the former two cases one easily checks that $x^2+bx+c$ does not divide $x^6+5x^3+8$, so we must have $c=2$. Calculating for this case the coefficient of $x^3$ gives $b^3-6b$, which must be$~5$, for which $b=-1$ is the unique integer solution. So one finds as unique factorisation of this kind $$ x^6+5x^3+8 = (x^2-x+2)(x^4+x^3-x^2+2x+4). $$ As we have shown that the left hand side has no other quadratic factors than the one shown on the right, the remaining quartic factor must be irreducible (unless it should be divisible by the quadratic factor, but it isn't).