Teacher2Teacher 
Q&A #104 
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From: Loyd <loydlin@aol.com> To: Teacher2Teacher Public Discussion Date: 2001092615:32:57 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(x1)=72 or x^2x72=0 has to be true. This factors and x is 9 and x1=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^2x12 and Exercise 2: 13x^245x  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^2x12) 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^2x12=6x^29x+8x12 (9x+8x=x) (6x^29x)+(8x12) Group as shown Factor each binomial: 3x(2x3) + 4(2x3) Note that (2x3) is a common factor. (3x4)(2x3). This is the desired result. Exercise 2: 13x^245x  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 752=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^252x+7x28 (13x^252x)+(7x28) Group as shown Factor each binomial: 13x(x4)+7(x4) = (13x+7)(x4) Desired result! We could have rearranged the equation (13x^2+7x)+(52x28) and the result would be: x(13x+7)4(13x+7) Remove common factor, (13x+7) and the answer is: (13x+7)(x4) 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 (x4) 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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