## Ask Professor Puzzler

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Dear Professor Puzzler, I read that if a polynomial equation has degree 2, there are two solutions, if it's degree three, there are three solutions, and so on. Is this true? Lucia

Well, Lucia,

It's true, and it's not true. It's true, with a couple of clarifications. Let's take a look to see a couple examples.

## Example One: A Perfect Square

x^{2}^{ }+ 6x + 9 = 0

How does this factor?

(x + 3)(x + 3) = 0, or (x + 3)^{2} = 0.

This gives us only one solution: x = -3. So the rule you read isn't exactly right in this case.

## Example Two: A "Complex" Polynomial

Consider this one: x^{2} + 9 = 0.

This can't be factored, and yet, if you know complex numbers, you can do this to it:

x^{2} = -9

x = 3i or -3i.

So we need to revise your rule a bit:

If you have a polynomial equation with degree n (n a positive integer), the number of solutions is n, as long as we agree to count repeated solutions once for each occurrence, and we allow for complex solutions.

So, for example, if you have an ugly polynomial like this:

x^{5}^{ }- 3x^{4} + 7x^{3} - 13x^{2} + 12x - 4 = 0

It can be factored as follows: (x - 1)^{3}(x + 2i)(x - 2i) = 0.

The solutions are: 2i, -2i, 1, 1, and 1, for a total of 5 solutions, which matches the degree of the polynomial.

By the way, if you're wondering how I factored x^{5}^{ }- 3x^{4} + 7x^{3} - 13x^{2} + 12x - 4, that's a post for another day!

Professor Puzzler