Question

15m^2+mn-2n^2 Factoring trinomials by using AC-Method.

Answer 1

what's the AC method?

Answer 2

It's for factoring trinomials..I am wondering if there is a different name

Answer 3

well here is my method to show you're teacher

for a quadratic

\[ax^2 + bx + c \]

multiply a and c

in your question its 15 and -2n^2 which is -30n^2

now find the factors of -30 that add to 1.....-5n and 6n

so you have the factors as

(15m - 5n)(15m + 6n)

now divide them by a... or 15

so you have

(15m - 5n)(15m + 6n)
--------------------
15

remove common factors from each binomial

5(3m - n) 3(5m + 2)
--------------------
15

which is really

15(3m -n)(4m +2)
-----------------
15

cancel the common factor of 15 and you have the solution.

Answer 4

Dont think thats right. if you multiply out your result to dont get the original expression.
i think its (5m+2n)(3m-n)

Answer 5

OKay but how did you do that?

Answer 6

the factors a right.

the second step would be

15m^2 + (6n - 5n) - 2n^2

Answer 7

then 15m^2 - 5nm + 6nm -2n^2

Answer 8

sorry 2nd step 15m^2 +(6n -5n)m - 2n^2

then 15m^2 - 5nm + 6nm - 2n^2

Answer 9

next 5m(3m -n) + 2n(3m -n)

Answer 10

you follow so far?

Answer 11

then (5m +2n)(3m -n)

Answer 12

so here is the method again

multiply a = 15 by c = -2n^2

you get -30n^2

find the factors that add to n

-5n, 6n

then the binomials need to be in the form

(am + factor 1)(am + factor 2)
-----------------------------
a

so you have

(15m -5n)(15m + 6n)
--------------------
15

factor each binomial

5(3m - n)3(5m + 2n)
-------------------
15

cancel the common factors

(3m - n)(5m _ 2n)

Answer 13

thanks campbell, i see why they call it the ac method.

Answer 14

this would be the possible AC method

and it works every time... provided the quadratic can be factored

Answer 15

damn, missed a + in the 2nd binomial

its (3m -n)(5m + 2n)