Assuming you mean: 7(x-2y)^2-25(x-2y)+12
7(x^2 - 4xy + 4y^2) - 25x + 50y + 12
7x^2 - 28xy + 28y^2 - 25x + 50y + 12
7x^2 - 25x - 28xy + 50y + 28y^2 + 12
.
How to factor that? Will it look something like this?
(x + y + 1)(x + y + 1)
x^2 + xy + x + xy + y^2 + y + x + y + 1
x^2 + 2x + xy + 2y + y^2 + 1
Yes. Something like (x + y + 1)(x + y + 1) will generate something like x^2 + 2x + xy + 2y + y^2 + 1
.
Back to the problem.
7x^2 - 25x - 28xy + 50y + 28y^2 + 12
If we're going to get a nice answer, we know the x's have to have 1s and 7s, so:
(we don't have to get a nice answer, but let's try)
(x + ?y + ?) (7x + ?y + ?)
(x - ?y + ?) (7x - ?y + ?)
(x - 2y + ?) (7x - 14y + ?)
Good. That will make the +28y^2 and the -28xy.
Now we need the numbers- the parts without an x or y. Those last numbers have to make +12, so our choices are: (1,12), (2,6), (3,4), (4,3), (6,2), (12,1), (-1,-12), (-2, -6), (-3, -4), (-4, -3), (-6, -2), (-12, -1).
Notice that we're going to end up with -25xy. That, along with our x and 7x, makes (3,4) look good. But -25xy is negative, so that makes (-3, -4) look good. Let's try that:
(x - 2y - 3) (7x - 14y - 4)
Check it:
7x^2 - 14xy - 4x -14xy + 28y^2 + 8y - 21x + 42y + 12
7x^2 - 25x - 28xy + 50y + 28y^2 + 12
Is that the same as 7x^2 - 25x - 28xy + 50y + 28y^2 + 12 ? Yes.
Answer: 7(x-2y)^2-25(x-2y)+12 factors to (x - 2y - 3) (7x - 14y - 4)