Q&A #104

Teachers' Lounge Discussion: Factoring trinomials

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From: Loyd

To: Teacher2Teacher Public Discussion
Date: 2001092616:32:57
Subject: Re: Factoring Trinomials Using the Regular Method

On 22 Jun 98 15:06:13 -0400 (EDT), Martha Kumler wrote:
>Here is a method to factor trinomials with leading coefficient >1
>   Steps                                   Example
>1. Multiply c by a, then drop           6x^2 - x - 12
>   the leading coefficient               x^2 - x - 72
>2. Factor the new trinomial             (x - 9)(x + 8)
>3. Replace the leading coefficient
>   in front of the x in each factor     (6x - 9)(6x + 8)
>4. Remove and discard all common 
>   factors within each parenthesis      (2x - 3)(3x + 4)
>   and you will have the factors 
>   of the original trinomial
>Note: If the original problem contains a common factor, factor it out
>first before following this procedure.  Ex:  6x^2 - 20x + 14
>                                           2(3x^2 - 10x + 7)

ax^2+bx+c (6x^2 -x -12)
It must be obvious that a and c are factors of ac  (6x12=72)
Also, The two factors we are looking for must add to -1.  Thus the
quadratic x(x-1)=72 or x^2-x-72=0  has to be true.  
This factors and x is 9 and x-1=8.  So, this part of your procedure is
valid and is almost the same procedure as most text books use up to
that point.  

If integral factors exist, then Your way is about the same as the
regular way until your step 3 which is hard to explain.  You need
reasons in math.

Try the regular way on two problems: Here is the one you presented and
another one a little more difficult:

Exercise 1: 6x^2-x-12         and Exercise 2: 13x^2-45x - 28.

Both of these follow the pattern, ax^2+bx+c and it is easy to see that
the product ac has factors a and c.  Likewise, 72 has factors 6 and 12
as well as several other pairs.
We will factor the expression in exercise 1:
Exercise 1:(6x^2-x-12)
 All we have to do is find a pair of factors whose sum = -1, the 
coefficient of the x term.  Immediately we see that -9 + 8 =-1.  We
simply rewrite the equation as:
6x^2-x-12=6x^2-9x+8x-12   (-9x+8x=-x)
(6x^2-9x)+(8x-12)        Group as shown
Factor each binomial:
3x(2x-3) + 4(2x-3)       Note that (2x-3) is a common factor. 

(3x-4)(2x-3).   This is the desired result.

Exercise 2: 13x^2-45x - 28   13x28=364.  We need factors of 364 whose
sum is -45.     We make a list of factors of 364
                                1  -364
                                2  -182
                                3      182 not divisible by 3
                                4  -91
                                5      182 not divisible by 5
                                6      364 not divisible by 6
                                7  -52  (Sum 7-52=-45)

Note:  To find these factors, I take half of one and double of the
other (DH) or 1/3 of one and 3 times the other etc until I find all
the necessary factors.  Divisibility rules help.

Using factors 7, -52 rewrite the equation as:
(13x^2-52x)+(7x-28)    Group as shown
Factor each binomial:
13x(x-4)+7(x-4) = (13x+7)(x-4)   Desired result!

We could have rearranged the equation (13x^2+7x)+(-52x-28) and the
result would be:
x(13x+7)-4(13x+7)      Remove common factor, (13x+7) and the answer

(13x+7)(x-4)  Which is the result we wanted.  This solution is not any
harder than the procedure you stated in your post.  

I have also noticed that students have trouble recognizing a binomial
factor such as (x-4) as a common factor.  They look for single
numerals or letters.  Recommend highlighting with color on your chalk
board to help understand this abstract idea. 

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