Factorise Completely 9y^2-81
factor out the 9:
9(y^2-9)
difference of squares:
9(y-3)(y+3)
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
To factorize the given expression, 9y^2 - 81, we first notice that both terms are perfect squares. The first term, 9y^2, can be written as (3y)^2, and the second term, 81, can be written as 9^2.
Now we can rewrite the expression as (3y)^2 - 9^2.
Using the difference of squares formula, we have:
(a^2 - b^2) = (a + b)(a - b),
where a = 3y and b = 9.
Substituting these values into the formula, we get:
(3y)^2 - 9^2 = (3y + 9)(3y - 9).
Therefore, the completely factored form of 9y^2 - 81 is (3y + 9)(3y - 9).