factor each trinominals

14m2+11mr-15r2

Why are you changing names on each post? If you want good help, you wouldn't do that.

factors of 14
14,1
7,2
factors of 15
15,1
3,5

what adds to 11 using the above numbers?

after a bit of mental reasoning..
7*3-2*5..

That is the thinking part. Put it together.

To factor the trinomial 14m^2 + 11mr - 15r^2, we want to find two binomials whose product equals this trinomial. We can use several methods to factor trinomials, such as factoring by grouping or using the quadratic formula, but in this case, we will use the AC method.

The AC method involves finding two numbers that multiply to give the product of the coefficient of the quadratic term (in this case, 14) and the constant term (in this case, -15), and also add up to the coefficient of the linear term (in this case, 11). Let's break down each step:

1. Multiply the coefficient of the quadratic term and the constant term:
14 * (-15) = -210

2. Find the two numbers that multiply to -210 and add up to 11. In this case, those numbers are 21 and -10:
21 * (-10) = -210
21 + (-10) = 11

Now, we will rewrite the linear term (11mr) using these two numbers:

14m^2 + 21mr - 10mr - 15r^2

Next, we will group the terms:

(14m^2 + 21mr) + (-10mr - 15r^2)

Now, we can factor by grouping. Factor out the greatest common factor from each group:

7m(2m + 3r) - 5r(2m + 3r)

As you can see, we now have a common binomial factor of (2m + 3r). Factoring it out, we get:

(2m + 3r)(7m - 5r)

Therefore, the factored form of the trinomial 14m^2 + 11mr - 15r^2 is (2m + 3r)(7m - 5r).