How do you factor this trinomial (if it's possible): 9m² - 12m + 4

Please show your work because I know the answer I just don't know how to get it.

9m^2-12m+4=9m^2-6m-6m+4 so using F-O-I-L in reverse we see that if our polynomial has to be in the form (?m+?)(?m+?) FOIL means the first two values are multiplied to get 9 and a good first choice might be 3*3. So now we have (3m+?)(3m+?) and we know that the last term in our polynomial is 4 and the last values in each set must multiply to equal 4 and a good first choice might be 2*2. So now we have (3m+2)(3m+2) which at first seems like it should work but we see it falls apart to create the -12m, but it does make a positive 12m so just a simple change to (3m-2)(3m-2) gives us the results we're looking for. So the final answer is (3m-2)^2. When you see that the first and last terms are squares a good first guess are the roots. Hope this helps.

(3m-2)(3m-2) = 0
(3m-2)^2 = 0

I use the grouping method where you find two factors of ax^2 ( which is 9m^2) and two factors of c (+4). You then cross multiply each factor of ax^2 with a factor of c, and sum the results and check whether the sum is equal to the coefficient of x.

Cross multiply

(3m)....[-2] = [-6m]
[3m]....(-2) = (-6m) +
------------------------------
...................-12m correct

If the coefficient of x was 13m

[9m]....(+4) = (4m)
(m)......[+1] = [9m] +
--------------------------------
....................13m

The contents of teh brackets is determined as the top and bottom row, so (9m+4) and (m+1)

I know this is a terrible explanation fo the grouping method as it is pretty difficult to explain so i suggest you research it.

USE FORMULA

a^2 -2ab + b^2 = ( a-b) ^2

9m^2 -12m + 4 = ( 3m)^2 - 2 (3m) (2) + (2)^2 = ( 3m -2) ^2 ANSWER

(3m-2)(3m-2)

(3m)^2 - 2*(3m)*2 + 2^2 = (3m-2)^2

9m² - 12m + 4 =(3m-2)^2

m^2 - 12m + 36

now solve it.