Factorise Completely 9y^2-81

Answer 1

factor out the 9:
9(y^2-9)
difference of squares:
9(y-3)(y+3)

Answer 2

Since 9 Can Divide In Both 9y^2 and 81 it is the lowest common multiple hence we write out the expression this way 9y^2-81= 9(y^2-9) & then we use difference of two squares to factorise what is in the brackets i.e a^2-b^2=(a+b)(a-b) so y^2-9=(y+3)(y-3) therefore 9[(y+3)(y-3)] is the simplified form and the 3^2[(y+3)(3-y)] is the factorised form