SOLUTION: Need help with finding the factor for the perfect square trinomial 64b to the second-112b+49 thanks

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Question 264016: Need help with finding the factor for the perfect square trinomial
64b to the second-112b+49 thanks

Answer by jim_thompson5910(28504) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)
In order to factor 64%2Ab%5E2-112%2Ab%2B49, first multiply the leading coefficient 64 and the last term 49 to get 3136. Now we need to ask ourselves: What two numbers multiply to 3136 and add to -112? Lets find out by listing all of the possible factors of 3136


Factors:

1,2,4,7,8,14,16,28,32,49,56,64,98,112,196,224,392,448,784,1568,3136,

-1,-2,-4,-7,-8,-14,-16,-28,-32,-49,-56,-64,-98,-112,-196,-224,-392,-448,-784,-1568,-3136, List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 3136.

1*3136=3136

2*1568=3136

4*784=3136

7*448=3136

8*392=3136

14*224=3136

16*196=3136

28*112=3136

32*98=3136

49*64=3136

56*56=3136

(-1)*(-3136)=3136

(-2)*(-1568)=3136

(-4)*(-784)=3136

(-7)*(-448)=3136

(-8)*(-392)=3136

(-14)*(-224)=3136

(-16)*(-196)=3136

(-28)*(-112)=3136

(-32)*(-98)=3136

(-49)*(-64)=3136

(-56)*(-56)=3136

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to -112? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -112

||||||||||||||||||||||
First Number|Second Number|Sum
1|3136|1+3136=3137
2|1568|2+1568=1570
4|784|4+784=788
7|448|7+448=455
8|392|8+392=400
14|224|14+224=238
16|196|16+196=212
28|112|28+112=140
32|98|32+98=130
49|64|49+64=113
56|56|56+56=112
-1|-3136|-1+(-3136)=-3137
-2|-1568|-2+(-1568)=-1570
-4|-784|-4+(-784)=-788
-7|-448|-7+(-448)=-455
-8|-392|-8+(-392)=-400
-14|-224|-14+(-224)=-238
-16|-196|-16+(-196)=-212
-28|-112|-28+(-112)=-140
-32|-98|-32+(-98)=-130
-49|-64|-49+(-64)=-113
-56|-56|-56+(-56)=-112


We can see from the table that -56 and -56 add to -112. So the two numbers that multiply to 3136 and add to -112 are: -56 and -56

So the original quadratic


64%2Ab%5E2-112%2Ab%2B49


breaks down to this (just replace -112%2Ab with the two numbers that multiply to 3136 and add to -112, which are: -56 and -56)


64%2Ab%5E2%2Bhighlight%28-56b-56b%29%2B49 Replace -112%2Ab with -56b-56b

Group the first two terms together and the last two terms together like this:

%2864%2Ab%5E2-56b%29%2B%28-56b%2B49%29

Factor a 8b out of the first group and factor a -7 out of the second group.


8b%288b-7%29%2B-7%288b-7%29


Now since we have a common term 8b-7 we can combine the two terms.


%288b-7%29%288b-7%29 Combine like terms.
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Answer:


So the quadratic 64%2Ab%5E2-112%2Ab%2B49 factors to %288b-7%29%288b-7%29

which can also be written as %288b-7%29%5E2 since the factors repeat themselves


Notice how %288b-7%29%288b-7%29 foils back to our original problem 64%2Ab%5E2-112%2Ab%2B49. This verifies our answer.