Question 190285: Factor
x^2+8xy+12y^2
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Looking at  we can see that the first term is  and the last term is  where the coefficients are 1 and 12 respectively.
Now multiply the first coefficient 1 and the last coefficient 12 to get 12. Now what two numbers multiply to 12 and add to the middle coefficient 8? Let's list all of the factors of 12:
Factors of 12:
1,2,3,4,6,12
-1,-2,-3,-4,-6,-12 ...List the negative factors as well. This will allow us to find all possible combinations
These factors pair up and multiply to 12
1*12
2*6
3*4
(-1)*(-12)
(-2)*(-6)
(-3)*(-4)
note: remember two negative numbers multiplied together make a positive number
Now which of these pairs add to 8? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 8
| First Number | Second Number | Sum | | 1 | 12 | 1+12=13 | | 2 | 6 | 2+6=8 | | 3 | 4 | 3+4=7 | | -1 | -12 | -1+(-12)=-13 | | -2 | -6 | -2+(-6)=-8 | | -3 | -4 | -3+(-4)=-7 |
From this list we can see that 2 and 6 add up to 8 and multiply to 12
Now looking at the expression , replace  with  (notice adds up to . So it is equivalent to )
Now let's factor  by grouping:
 Group like terms
 Factor out the GCF of  out of the first group. Factor out the GCF of  out of the second group
 Since we have a common term of  , we can combine like terms
So factors to
So this also means that factors to (since is equivalent to )
------------------------------------------------------------
Answer:
So factors to
|
|
|
