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A Special Case: Factoring "In Pairs" (page 2 of 2)

Sections: Simple factoring, Factoring "in pairs"


There is one special case that you may or may not need, depending upon how your book is structured and how your instructor intends to teach factoring quadratics. I call it "factoring in pairs".

  • Factor xy – 5y – 2x + 10.

    Is there anything that factors out of all four terms? No. When you have four terms, and nothing factors out of all of them, think of "factoring in pairs". That is, split the expression into two pairs of terms, and factor the pairs separately.

      xy – 5y – 2x + 10

    What can I factor out of the first pair? I can take out a "y":

      xy – 5y – 2x + 10

        = y(x – 5) – 2x + 10

    What can I factor out of the last pair? I can take out a "–2":

      xy – 5y – 2x + 10

        = y(x – 5) – 2x + 10

        = y(x – 5) – 2(x – 5)

    (I took out a –2, rather than a 2, because the leading sign on the pair was a "minus". And I got a "–5" in the result because, when I divided the positive 10 by the negative 2, the result was a negative 5. Be careful with your signs!)

    Now I do have a common factor, so I can proceed as usual:

      xy – 5y – 2x + 10

        = y(x – 5) – 2(x – 5)

        = (x – 5)(y – 2)

Factoring "in pairs" is most commonly used to introduce factoring quadratics. So you may see problems that look like this:

  • Factor x2 + 4xx – 4.

    This is four terms with no common factor, so factor "in pairs":

      x2 + 4xx – 4

        = x(x + 4) – 1(x + 4)

        = (x + 4)(x – 1)

Note how, in the second line, I factored a "1" out.  If "nothing" factors out, a "1" factors out.

  • Factor x2 – 4x + 6x – 24.

    Factor "in pairs": Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

      x2 – 4x + 6x – 24

        = x(x – 4) + 6(x – 4)

        = (x – 4)(x + 6)


If you will be using "factoring in pairs" for factoring quadratics (which is not the method I use), it will be called "factoring by grouping", and it will work like this:

  • Factor x2 – 5x – 6.

    Find factors of –6 that add up to –5: use the number –6 and +1, because (–6)(+1) = –6,
    and
    (–6) + (+1) = –5. Split the middle "–5x" term into "–6x" and "+1x", and then factor in pairs:

      x2 – 5x – 6

        = x2 – 6x + 1x – 6

        = x(x – 6) + 1(x – 6)

        = (x – 6)(x + 1)

  • Factor 6x2 13x + 6.

    Find factors of (6)(6) = 36 that add up to –13: use the numbers –9 and –4, because (–9)(–4) = 36, and (–9) + (–4) = –13. Split the middle "–13x" term into "–9x" and "–4x", and factor in pairs:

      6x2 13x + 6

        = 6x2 9x – 4x + 6

        = 3x(2x – 3) – 2(2x – 3)

        = (2x – 3)(3x – 2)

For further explanations and examples of this last bit, go to the lesson on factoring quadratics.

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Cite this article as:

Stapel, Elizabeth. "A Special Case: Factoring 'In Pairs'." Purplemath. Available from
    http://www.purplemath.com/modules/simpfact2.htm. Accessed
 

 

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