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

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

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?

Yess

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?

So far so good

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 \[11amn+cmn=25mn\]

im writing this down just a sec :)

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

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

exactly! so what is the factorization in this case?

Two and Three?

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

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

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

beautiful :D great job!

Thank you so much :) I appreciate it

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

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.

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

Happy to help good luck!

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