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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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