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Factoring Quadratics: The Simple Case (page 1 of 4)

Sections: The simple case, The hard case, The weird case


A "quadratic" is a polynomial that looks like "ax2 + bx + c", where "a", "b", and "c" are just numbers.

For the easy case of factoring, you will find two numbers that will not only multiply to equal the constant term "c", but also add up to equal "b", the coefficient on the x-term. For instance:

  • Factor x2 + 5x + 6.
  • I 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 I'll use 2 and 3. I 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 I'll draw my parentheses, with an "x" in the front of each:

      (x       )(x      )

    Then I'll write in the two numbers that I 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.

  
Note that you can always check your work by multiplying back to get the original answer. In this case:

  

 (x + 2)(x + 3) = x^2 + 5x + 6

 

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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, in addition to finding the factors of 6 that add to 5, you would have had to do these additional steps:

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

          = (x + 3)(x + 2)

You get the same answer as by the previous method, but I think it's easier to just fill in the parentheses.

  • 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 above 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 I multiplied to a positive six, then the factors must have the same sign. (Remember that two negatives multiply to a positive.) Since I'm 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 I 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 I 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 1999-2011 All Rights Reserved

  • Factor x2 – 7x + 6.
  • In this case, I am multiplying to a positive six, so the factors are either both positive or both negative. I am adding to a negative seven, so the factors are both negative. The factors of 6 that add up to 7 are 1 and 6, so I will 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 I am multiplying to a negative six, I need factors of opposite signs; that is, one factor will be positive and the other will be negative. The larger factor (in absolute value) will get the "plus" sign, because I am adding to a positive 1. Since these opposite-signed numbers will be adding to 1, I need the two factors to be one unit apart. The factor pairs for six are 1 and 6, and 2 and 3. The second pair are one apart, so I 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 x2x – 6.
  • This looks just like the previous case, except that now the middle term is negative. I still want factors with opposite signs, and I still want factors that are one apart, but this time the larger factor gets the "minus" sign:

      x2x – 6 = (x – 3)(x + 2)

  • Factor x2 – 5x – 6.
  • In this case, I still want factors of opposite signs, but now I want them to be five units 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 I'll 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, I'll be needing a positive and a negative number such that, when I multiply them together, I get 6, but when I add them, I get 7. The factor pairs for 6 are 1 and 6, and 2 and 3. You may think that I should use 1 and 6, but —

    One of the factors has to be negative in order to multiply to get a "minus" six! Trying the first factor pair of 1 and 6, the sum would be either (–1) + 6 = 5 or else 1 + (–6) = –5. And the other factor pair, 2 and 3, won't work, either, because (–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 I couldn't find integers that would work).

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Cite this article as:

Stapel, Elizabeth. "Factoring Quadratics: The Simple Case." Purplemath. Available from
    http://www.purplemath.com/modules/factquad.htm. Accessed
 

 

 

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