here is the equation; 6m^2+11mn+3n^2
usually, a and b are perfect squares, allowing me to easily factor [(a+b)^2]
now that 6 and 3 aren't perfect squares, how do I factor this?
Oh and the answer is; =(3n+2m)(n+3m)
Note: Please use the decomposition method so I can understand how to do this the way I was taught, please
It appears that the "decomposition method" is another name for the "a-b-c method". (
here) However, that method does not require that the quadratic expression (not "equation", since it does not contain an "equals" sign) to be a perfect-square trinomial.
To learn how the method works for
all factorizable quadratic expressions (factorizable over the rationals, anyway), try
here.
Working from the explanation in the linked lesson (and noting that your quadratic is an example of "the hard case"), we see the following:
. . . . .a = 6
. . . . .b = 11
. . . . .c = 3
. . . . .a*c = 18
We're multiplying to a positive, so the factors of 18 must have the same sign. We're adding to a positive, so that sign must be "plus". The factor pairs for 18 are:
. . . . .1 and 18
. . . . .2 and 9
. . . . .3 and 6
Which pair adds to 11? This is the pair you will need to use. Plug this into the "box" (which is a means of keeping signs straight when using the decomposition method), and do the factorization. Or, if you prefer, use the two factors in the correct pair to "decompose" the middle term in the original quadratic. Then factor "in pairs".