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Mathematics
OpenStudy (anonymous):

Factor: 6m^2+25mn+11n^2 (please help, I have no idea what i'm doing)

OpenStudy (turingtest):

there may be better ways to do this, but mine relies largely on trial and error...

OpenStudy (turingtest):

the first thing is to try to figure out what you do know about how the equation will factor it should look like\[(am+bn)(cm+dn)\]where a,b,c, and d are the numbers we need to find do you agree so far?

OpenStudy (anonymous):

Yess

OpenStudy (turingtest):

now what do we know about a, b, c, and d? 1)they are all positive, because all the terms in our original expression are positive 2)because 11 is prime, b and d must be 1 and 11 (which is which does not matter yet) still with me?

OpenStudy (anonymous):

So far so good

OpenStudy (turingtest):

since it makes no difference yet, let b=1 and d=11 so we know we will have something of the form\[(am+n)(cn+11n)\]so now we only have a few choices to try for what a and c must be, since they must multiply to 6 so we only have to check a=1,c=6 a=2,c=3 a=3,c=2 a=6,c=1 and whichever combination gives us the correct middle coefficient of \[11amn+cmn=25mn\]

OpenStudy (anonymous):

im writing this down just a sec :)

OpenStudy (turingtest):

just looking at the coefficients (or you could say 'dividing this by mn) the requirement is that\[11a+c=25\]so try the four possibilities above and you will see which is correct

OpenStudy (anonymous):

It's the second choice = 11(2)+3=25

OpenStudy (turingtest):

exactly! so what is the factorization in this case?

OpenStudy (anonymous):

Two and Three?

OpenStudy (turingtest):

but we agreed that the factored form is\[(am+bn)(cm+dn)\]so what are a, b, c, and d in this case? plugging in those numbers into the form above is the answer we are looking for

OpenStudy (anonymous):

a= 2 b= 1 c= 3 d= 11

OpenStudy (anonymous):

so the answer is (2m+1n)(3m+11n) ? Correct?

OpenStudy (turingtest):

beautiful :D great job!

OpenStudy (anonymous):

Thank you so much :) I appreciate it

OpenStudy (turingtest):

you're welcome I will say one final thing that is a bit more advanced, but completes the ideas above...

OpenStudy (turingtest):

The thing to remember are the requirements on factoring something of the form\[Ax^2+Bxy+Cy^2=(ax+by)(cx+dy)\]By checking what we get by FOIL-ing out the parentheses, we know that \(A=ac\), \(B=ad+bc,\) and \(C=bd\) Those are the facts I used to narrow down the possibilities of \(a,b,c, \text{ and } d\) The coefficient \(B\) is called the \(cross-product\), and above you will notice that the 4 choices we got for a and c were decided between by checking the cross-product. That is the main trick I use in this kind of factoring.

OpenStudy (anonymous):

this method is alot easier than the one in my text book,, i'll save this to my notes!

OpenStudy (turingtest):

Happy to help good luck!

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