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Q&A #18119


"Bottoms up" as a method to factor trinomials

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From: Nancy Kitt <nkitt@math.iupui.edu>
To: Teacher2Teacher Service
Date: Dec 29, 2006 at 21:33:53
Subject: "Bottoms up" as a method to factor trinomials

One of my former students showed me the following method to factor trinomials. I want to know HOW and WHY this method works. 3x^2 + 14x + 8 Multiply AC, that is 3 x 8 = 24 Now look at B = 14. We are looking for two numbers multiplied together to give 24 and added to give 14. The numbers will be +12 and +2. (x + 12)(x + 2)--- put the two factors 12 and 2 inside the parentheses, but put x as the first term in both parentheses. Now, since A was 3, divide the two factors 12 and 2 by 3 (x + 12/3) (x + 2/3) 12 will divide by 3 giving 4. 2 does not divide by 3. Therefore, multiply the x by 3, giving the final factorization of (x + 4 (3x + 2). Another example: 12x^2 + 13x - 35 A times C gives 420 and B = 13. Look for two numbers multiplied to give 420 and added to give 13. The factors are +28 and -15. (x + 28)(x - 15)-- put the two factors +28 and -15 inside the parentheses, but put x as the first term in both parentheses. Now, since A was 12, divide the two factors, +28 and -15 by 12. (x + 28/12)(x - 15/12) Reduce +28/12 to +7/3 and reduce -15/12 to -5/4. Since neither reduce to an integer, in the first parentheses, the last term will be +7 (the numerator of the reduced fraction), but multiply x by 3 (the denominator of the reduced fraction). In the second parentheses, the last term will be -5 (the numerator of the reduced fraction), but multiply x by 4 (the denominator of the reduced fraction). Therefore, the factors are (3x + 7) and (4x - 5). This is the COOLEST method I've ever seen. However, I have NO CLUE HOW or WHY it works!!!!!! I want to use this method this semester, and I'd like to have an idea why it works?????

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