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Lesson pages:  The simple case,  The hard case,  The weird case

The simple case

A "quadratic" is a polynomial that looks like "ax² + bx + c", where "a", "b", and "c" are just numbers. For the easy case, you will find two numbers that multiply to the constant term "c", and add to "b", the coefficient on the x-term. For instance:

Factor x² + 5x + 6.

We need to find factors of 6 that add up to 5. Since 6 can be written as the product of 2 and 3, and since 2 + 3 = 5, then we'll use 2 and 3. Now, you know from multiplying polynomials that this quadratic is formed from multiplying two factors of the form "(x + m)(x + n)", for some numbers m and n. So draw your parentheses, with an "x" in the front of each:

(x       )(x      )

Then write in the two numbers we found above:

(x + 2)(x + 3)

This is the answer:  x² + 5x + 6 = (x + 2)(x + 3)

This is how all of the "easy" quadratics will work: you will find factors of the constant term that add up to the middle term, and use these factors to fill in your parentheses.

Your text or teacher may refer to factoring "by grouping", which is covered in the lesson on simple factoring. In the "easy" case of factoring, using "grouping" just gives you some extra work. For instance, in the above problem, you would still have had to find the factors of 6 that add to 5. But instead of just filling in the parentheses, you would have done these steps:

x² + 5x + 6 = x² + 3x + 2x + 6 = (x² + 3x) + (2x + 6) = x(x + 3) + 2(x + 3) = (x + 3)(x + 2)

You get the same answer, but I think it's easier to just fill in the parentheses. Here are some more examples:

Factor x² + 7x + 6.

The constant term is 6, which can be written as the product of 2 and 3 or of 1 and 6.  But 2 + 3 = 5, so 2 and 3 are not the numbers I need in this case. On the other hand, 1 + 6 = 7, so I'll use 1 and 6:

x² + 7x + 6 = (x + 1)(x + 6)

Note that the order doesn't matter in multiplication, so the answer could equally correctly be written as "(x + 6)(x + 1)".

Factor x² – 5x + 6.

The constant term is 6, but the middle coefficient this time is negative. Since we multiplied to a positive six, then the factors must have the same sign. (Remember that two negatives multiply to a positive.) Since we're adding to a negative (–5), then both factors must be negative. So rather than using 2 and 3, as in the first example, this time we will use –2 and –3:

x² – 5x + 6 = (x – 2)(x – 3)

Note that you can use clues from the signs to determine which factors to use, as we did in this last example above:

If c is positive, then the factors you're looking for are either both positive or else both negative.
If b is positive, then the factors are positive
If b is negative, then the factors are negative.
In either case, you're looking for factors that add to b.

• If c is negative, then the factors you're looking for are of alternating signs; that is, one is negative and one is positive.
  If b is positive, then the larger factor is positive.
  If b is negative, then the larger factor is negative.
  In either case, you're looking for factors that are b units apart.

Let's try another one...   Copyright © Elizabeth Stapel 2000-2004 All Rights Reserved

Factor x² – 7x + 6.

In this case, you are multiplying to a positive six, so the factors are either both positive or both negative. You are adding to a negative seven, so they are both negative. Factors of 6 that add to 7 are 1 and 6, so use –1 and –6:

x² – 7x + 6 = (x – 1)(x – 6)

So far, "c" has always been positive. What if c is negative?

Factor x² + x – 6.

Since you are multiplying to a negative six, you need factors of opposite signs; that is, one will be positive and the other will be negative. The larger one will have a "plus" sign, however, because you are adding to a positive 1. And you need the factors to be one apart. The factor pairs for six are 1 and 6, and 2 and 3. This second pair are one apart, so you want to use 2 and 3, with the 3 getting the "plus" sign (so the 2 gets the "minus" sign). 

x² + x – 6 = (x – 2)(x + 3).

Factor x² – x – 6.

This looks just like the previous case, except that now the middle term is negative. You still want factors with opposite signs, and you still want factors that are one apart, but this time the larger factor gets the "minus" sign:

x² – x – 6 = (x – 3)(x + 2)

Factor x² – 5x – 6.

In this case, you still want factors of opposite signs, but now you want them to be five apart (and the larger factor will get the "minus" sign). The factor pairs for six are 1 and 6, and 2 and 3. The first pair are five apart, so use the numbers +1 and –6:

x² – 5x – 6 = (x – 6)(x + 1)

There is one special case, by the way, for factoring. Back when you were factoring plain old numbers, there were some numbers that didn't factor, such as 5 or 13. Recall that they are called "prime" numbers. The terminology is the same for polynomials:

Factor x² + 7x – 6.

Since the constant term is negative, you'll be needing a positive and a negative number such that, when you multiply them together, you get 6, but when you add them, you get 7. The factor pairs for 6 are 1 and 6, and 2 and 3. You may think that you should use 1 and 6, but--- one of them has to be negative in order to multiply to get a minus six!  Then the sum would be either (–1) + 6 = 5 or else 1 + (–6) = –5.  And 2 and 3 won't work, either:  (–2) + 3 = 1, and 2 + (–3) = –1.

In other words, there is no pair of factors of –6 that will add to +7. And if something isn't factorable? It's prime. Then x² + 7x – 6 is "prime", or "unfactorable over the integers" (because we couldn't find integers that would work).

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