
A Special Case: Factoring "In Pairs" (page 3 of 3) Sections: Simple factoring, Factoring "in pairs" There is one special case for factoring 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".
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". To factor "in pairs", I split the expression into two pairs of terms, and then 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 second 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 positive10 by the negative2, the result was a negative5. Be careful with your signs!) Now that I do have a common factor, 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 exercises that look like this:
This polynomial has four terms with no factor common to all four, so I'll try to factor "in pairs": x^{2} + 4x – x – 4 = x(x + 4) – 1(x + 4) = (x + 4)(x – 1) In the second line above, I factored a "1" out. Why? If "nothing" factors out, a "1" factors out.
I'll try to factor "in pairs": Copyright © Elizabeth Stapel 20022011 All Rights Reserved x^{2} – 4x + 6x – 24 = x(x – 4) + 6(x – 4) = (x – 4)(x + 6) You can use the Mathway widget below to practice factoring a polynomial by grouping. Try the entered exercise, or type in your own exercise. Then click "Answer" to compare your answer to Mathway's. (Or skip the widget and continue with the lesson.)
(Clicking on "View Steps" on the widget's answer screen will take you to the Mathway site, where you can register for a free sevenday trial of the software.) If you will be using "factoring in pairs" for factoring quadratics (which is not the method I use), your book will refer to it as something like "factoring by grouping", and it will work like this:
First, I have to find factors of the last term, –6, that add up to the numerical coefficient of the middle term, –5. I'll use the number –6 and +1, because (–6)(+1) = –6, and (–6) + (+1) = –5. Using these numbers, I'll split the middle "–5x" term into the two terms "–6x" and "+1x". This will then allow me to factor in pairs: x^{2} – 5x – 6 = x^{2} – 6x + 1x – 6 = x(x – 6) + 1(x – 6) = (x – 6)(x + 1)
This is a bit more complicated, because the leading coefficient (the number on the x^{2}) is not a simple "1". But I can still factor the polynomial. First, I need to find factors of (6)(6) = 36 that add up to –13. I'll use the numbers –9 and –4, because (–9)(–4) = 36, and (–9) + (–4) = –13. Then I can split the middle "–13x" term into the two terms "–9x" and "–4x", and then factor in pairs: 6x^{2} – 13x + 6 = 6x^{2} – 9x – 4x + 6 = 3x(2x – 3) – 2(2x – 3) = (2x – 3)(3x – 2) For a complete explanation of these last two examples (whose method may look somewhat "magical" at the moment), please study my lesson on factoring quadratics. The "simple" and then the "hard" case pages should completely clarify the topic. << Previous Top  1  2  3  Return to Index


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