basic quadratics help for equation 6m^2 + 11mn + 3n^2

Drazil

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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
Thanks!
 
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
Thanks!

The factored form must look like this (am+bn)(cm+dn)

a*c = 6, a = 6 and c = 1 or a = 3 and c = 2 or a = 1 and c = 6 or a = 2 and c = 3. There are no other choices.

b*d = 3, b = 1 and d = 3 or b = 3 and d = 1. There are no other choices.

a*d + b*c = 11

The hunt is afoot!
 
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
Thanks!
The way you are being taught is idiotic. Many quadratics with real coefficients cannot be factored over the real numbers, and the majority of those that can be factored over the real numbers cannot be factored over the integers. So you are being taught a method that usually will not work.
 
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
Thanks!

Perfect square trinomials are a very special case, (a+b)^2 = a^2 + 2ab + b^2. If the first and last terms of a trinomial are squares, it may have this form, but even then it may not. It is not "usual".

There are various ways to factor a trinomial in general; but I don't know which method you are calling "the decomposition method". Can you show us an example you were given, so we can see how you were taught?

By the way, it is not an equation; it is a polynomial expression. Confusing those words leads to many misunderstandings.
 
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". ;)
 
Lazy me prefers using the quadratic equation, which leads to:
a = -3b/2 or -b/3
 
Lazy me prefers using the quadratic equation, which leads to:
a = -3b/2 or -b/3
Denis, I agree.

With the kids I tutor, I tell them to give factoring quadratics 60 seconds. If that works, great. If not, use the quadratic formula. If the exercise says to use factoring and you don't see it, use the quadratic formula and fit it into an answer in factoring form.

As tkhuny says, correct answers don't care how you find them, and neither does real life.
 
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