`ax^(2)+(4a^(2)-3b)x-12ab`

Solve by factorization: a x^2+(4a^2-3b)x-12 a b=0

If 4a^(2)+9b^(2)-c^(2)-12ab=0 then family of straight lines ax+by+c=0 are concurrent at P and Q then values of P & Q are

Factorization by making a perfect square: 4a^(2)+12ab+9b^(2)-8a-12b

If 4a^(2)+9b^(2)-c^(2)+12ab=0 then the family of straight lines ax+by+c=0 is concurrent at : (A)(-3,2) or (2,3)(B)(-2,3) or (2,-3)(C)(3,2) or (-3,-2)(D)(2,3) or (-2,-3)

If 4a^(2)+9b^(2)-c^(2)+12ab=0 , then the set of lines ax + by+c = 0 pass through the fixed point

If 4a^(2)+9b^(2)-c^(2)+12ab=0 , then the set of lines ax+by+c=0 pass through the fixed point

Statement I: The points (a,0),(0,b) and (1,1) will be collinear if 1/a+1/b=1 Statement II: If 4a^(2)+9b^(2)-c^(2)+12ab=0 , then the family of lines ax+by+c=0 is either concurrent at (2,3) or at (-2,-3). Then which of the followng is true

Factorize each of the following expressions: 4a^(2)+12ab+9b^(2)-8a-12ba^(2)+b^(2)-2(ab-ac+bc)