Factor the trinomial. 12x^2+20x-25

Explain this to me so I can do it on my own please! I'd appreciate it!!

If you are able to factor this guy it would be in the form (Ax+B)(Cx-D)
where the A and C are factors for 12
and B and D are factors for 25
the factors for 25 are : 1,5,25 so pairs would be(B=1,D=25)(B=5, D=5) or (B=25, D=1)
let try B=5 and D=5
so then the factors would look like this (Ax+5)(Cx-5) = (AC)x² +(5C-5A)x -25

looking at the middle term
(5C-5A)x has to equal 20x or (5C-5A)=20 or 5(C-A)=20 or C-A=4 so now look at the factors for 12 and see if any of them will make this equation true
factors of 12
(A,C)=(1,12)(2,6)(3,4)(4,3)(6,2)(12,1)
to get C-A to equal 4 you will need (2,6) because 6-2=4
so A=2 and C=6
the factors are (2x+5)(6x-5)

hope this helps. try it on a few. if you did not choose the right values for B and D then you will not get values that will work for A and C to give you the middle term. then you have to try another pair for B and D.
dnadan1

Product sum method first:
P = -25*12x^2= -300x^2
S = 20x
now lets find two terms that satisfy both above terms (multiply to -300x^2 and add to 20x)
P = 30x * (-10x)
S = 30x + (-10x)

since the a in this trinomial is not 1 we have to use the decomposition method
12x^2 + 20x - 25
= 12x^2 +30x -10x -25
= (12x^2 + 30x) + (-10x - 25)
= 6x(2x +5) -5(2x + 5)
= (6x - 5)(2x + 5)

(2x+5)(6x-5)
see step by step solution:
http://symbolab.com/solutions/factor_qua…

when you have nominal in the form of ax^2 + bx + c, the find the factors of the product ac the sub to b, then factor via grouping.

ac = 12 * -25 = -300

Factorise using the FOIL method: (ax+b)(cx+d) = (2x+5)(6x-5).