Trial and Error Examples

Factor the polynomial x² + 8x + 9, if possible.

We sort of gave away the answer with the wording of this problem. Oops. We have a problem with that. If you haven't seen The Usual Suspects, you should steer clear of us. We will definitely blow the ending for you.

It isn't possible to factor this polynomial, but why not? If we could factor the polynomial as (x + m)(x + n), we would need mn = 9 and m + n = 8.

There aren't two integers that multiply together to give 9 and add together to give 8. We tried everything; we even took out an ad in the paper, but no luck: 1 and 9 don't work; 3 and 3 don't work; -1 and -9 don't work; -3 and -3 don't work. There aren't any pairs of numbers left to try that multiply to give 9, so we can't factor the polynomial. Now we're out $25 for that ad.

Factor the polynomial 2x³ – 8x² + 6x.

This polynomial isn't quadratic. However, all the terms have a common factor of 2x. Let's take that out and see what happens. First:

(2x)(x² – 4x + 3)

The second factor is quadratic. Can we factor it further? We need two numbers whose product is 3 and whose sum is -4. Hey, we happen to know the two numbers that will fit the bill: -3 and -1 work, so we can factor the original polynomial like so:

(2x)(x – 3)(x – 1)

Remember, you can always check your answer to a factoring problem by multiplying out the factors. The product of the factors should be the original polynomial. If it isn't, you've gone wrong somewhere. Maybe you took a wrong turn at Algebraquerque.