Factories `y^3-3y^2-9y-5​`

Factorise : 4x^2-9y^2-2x-3y

Factories: 9y^(2) - 12y +4

Factorise 25x^(2) - 30xy + 9y^(2) . The following steps are involved in solving the above problem . Arrange them in sequential order . (A) (5x - 3y)^(2) " " [ because a^(2) - 2b + b^(2) = (a-b)^(2)] (B) (5x)^(2) - 30xy + (3y)^(2) = (5x)^(2) - 2(5x)(3y) + (3y)^(2) (C) (5x - 3y) (5x - 3y)

The perimeter of a triangle is 11y^(2)-9y+13 and two sides are 4y^(2)-3y-7 and 5y^(2)-4y+6 then find the third side of the triangle

(y-3)^(2)=y^(2)-9

Factorize :7(2x+5)+3(2x+5)(x+2)y+(x+2)x5a(2x+3y)-2b(2x+3y)8(5x+9y)^(2)+12(5x+9y)

If the lines 2x^(2)-5xy+3y^(2)+8x-9y+6=0 and kx-y-5=0 are concurrent, then k=

The point of intersection of the lines 2x^(2)-5xy+3y^(2)+8x-9y+6=0 is

Solve: (9-3y)/(1-9y)=8/5