This question comes from one of my Pre-Cal students, who wants to know if there's a nice rule like the "sum and product" factoring rule, that works when the roots are complex numbers.

The sum-and-product method he's referring to is this:

If we have a quadratic x² + bx + c, and we can find two numbers (let's call them m and n) that add to b and multiply to c, then the quadratic factors into (x + m)(x + n). Here's an example:

Factor x² + 13x + 36.

In order to factor this, I look for two numbers that add to 13, and multiply to 36. Can you find them? I usually start by asking, "What are two numbers that multiply to 36?" The first pair I think of is 6 x 6. But 6 + 6 isn't 13, so that doesn't work. Next I think of 9 x 4. Aha! 9 + 4 is 13. So, since 9 x 4 = 36 and 9 x 4 = 13, I conclude that:

x² + 13x + 36 = (x + 4)(x + 9).

Okay, so now the question is, what if the quadratic factors with complex numbers? Is there a nice way of doing this factoring rule?

Let's start with an example:

x² - 4x + 5

If I knew that the factorization of this resulted in complex numbers, I would remember another rule, that complex roots come in pairs: if one root is a + bi, then the other is a - bi.

So, we look for two numbers such that:

(a + bi) + (a - bi) = -4
(a + bi)(a - bi) = 5

Simplifying these gives:

2a = -4, or a = -2
a² + b² = 5
4 + b² = 5
b² = 1
b = ⁺/_- 1

Therefore, this factors as follows:

x² - 4x + 5 = (x - 2 - i)(x - 2 + i) Multiply this out, and sure enough, it's true.

Let's try generalizing this a bit. Suppose the quadratic was x² + mx + n, and we wanted to find complex roots with the sum and product method.

(a + bi) + (a - bi) = m
(a + bi)(a - bi) = n

2a = m
a² + b² = n

a = m/2

m²/4 + b² = n
b = SQR(n - m²/4)

So let's take another example, and put our rule into practice.

x² + 6x + 13

a is half of 6, or 3.

b is SQR{13 - 6²/4) = SQR{13 - 9} = 2.

Therefore x² + 6x + 13 = (x +3 + 2i)(x + 3 - 2i).

This isn't too bad, but it presumes that the leading coefficient of the quadratic is 1; things might get a bit dicier if it wasn't. And, of course, if you're paying close attention, the result we obtain is actually just a slightly re-arranged version of the quadratic formula as it would appear in the special case of the leading coefficient being one!