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A problem that is an example of this is $x^2+16$. I have to know how to factor this over the set of complex numbers. How do you do that? I used to know it's just been a long time.

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3 Answers 3

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Notice that $16 = -(4i)^2$, so $x^2 + 16$ is a difference of two squares and can be written as $(x - 4i)(x + 4i)$.

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Ah... thanks now I remember. –  chromedude Sep 25 '10 at 20:55
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@chromedule. And, just if you don't notice the smart remark of muad, you can just put $x^2 +16 = 0 \ \Leftrightarrow \ x = \sqrt{-16}$ and get the same result using the fact that for a polynomial $P(x)$ is the same to say that $\alpha$ is a root of $P(x)$ that $x-\alpha$ is a factor of $P(x)$. –  a.r. Sep 25 '10 at 22:05

In general, it is true that every polynomial splits (i.e., factors into linear factors) over the complex number field. However, finding this factorization for polynomials of degree up to four is manageable, and factoring something of degree higher than 4 is impossible (in the sense that there does not exist, for a general polynomial of degree 5 or higher, an algorithm for finding its roots). Well, actually, this isn't entirely true. I should say that there is no formula for the roots of a general quintic expressed by radicals. There are other methods: for instance, the so called ultraradicals, approximations, etc.

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The easiest way to factor polynomials is by finding roots: if $p(a)=0$ then $p(x)=(x-a)q(x)$ with $q(x)$ being a polynomial of smaller degree than $p(x)$ and you can repeat the process by finding a root of $q(x)$. So if you have a root $a$, you simply compute $q(x)=p(x)/(x-a)$ and continue recursively. This works over any field - the power of the Complex numbers stems from the fact that this process always works in the complex number - every polynomial has a root.

Still, finding such a root is not always an easy task. For polynomials of degree 2 it is trivial using the well known formula $z_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$ which gives the roots of the polynomial $az^2+bz+c$. In simpler cases, such as $x^2+16$ you gave above, one can solve directly ($x^2=-16$ and then taking roots you find $\pm 4i$). However, for higher degrees the task becomes more difficult. There is a formula for polynomials of degree 3 and 4 but it is more complex; and for degree 5 you don't have a simple formula anymore (with "simple" meaning "involving only the four operations of arithmetic and taking roots").

For those cases, it is simpler to find an approximate solution and use the division method described above. A typical method for finding approximate roots of polynomials is the Newton-Rapson method.

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