Question

Factorise:- 7 (2x-y) 2 - 25 (x-2y) + 12

Answer 1

The question is not clear. It should be like this 7(2x-y)^2 - 25(x-2y) + 12
Solution:-
7(2x-y)^2 - 25(x-2y) + 12
[7*{(2x-y)^2}] - 25*(x-2y) + 12
{7(4x^2 - 4xy +y^2)} - 25x + 50y + 12
28x^2 - 28xy + 7y^2 - 25x +50y +12
28x^2 - 28xy - 25x + 7y^2 + 50y +12
These have no common factor so can't be factorize further.

Answer 2

P(x,y) = 7 (2 x - y)² - 25 (x - 2 y) + 12
     = 28 x² + 7 y² - 28 x y - 25 x - 50 y + 12
     = (a x + b y + c) ( 28/a x + 7/b y + 12/c)        Let it be
     = 28 x² + 7 y² + 12 + [28b/a + 7a/b] xy + [7c/b+12 b/c] y + [28c/a+12a/c]x
compare coefficients and let  
      b/a = m          c/b = n         c/a = p

so:   28 m + 7/m = -28 
           4 m² + 4 m + 1 = 0     =>  m = -1/2   =>  a = - 2 b

and   7 n + 12/ n = -50
            7 n² + 50 n + 12 = 0   =>  n = [-25 +- √[541] / 7
           =>  c = n b

and   28 p + 12 / p = -25
          28 p² + 25 p + 12 = 0
       => Discriminant:  625 - 4*12*28 <0

so there is no solution such that a, b, c are consistent.

There are no two linear factors of degree 1  for the given expression in x and y.