Factorise the following
1.(x - 2y)^2 -16/49
2. 3/64 - 27y^2
I will give you a hint... the first one is called a "difference of squares"
I will do an easier difference of squares and see if you remember...
x^2 - 9 = (x - 3)(x + 3) and remember the sign have to be opposites so when you do the distributive property you don't get a third term : )
Does this hint help?
Form: a^2 - b^2 = (a+b)(a - b).
1. (x-2y)^2 - 16/49 = ((x-2y) + 4/7)((x-2y) - 4/7)
To factorize the given expressions, we'll try to express them as a product of two binomial expressions.
1. (x - 2y)^2 - 16/49:
To factorize this expression, let's simplify it first.
(x - 2y)^2 - 16/49 can be written as [(x - 2y) - 4/7]^2.
Now, this expression is in the form (a - b)^2, which can be factored as (a - b)^2 = (a - b)(a - b).
So, (x - 2y)^2 - 16/49 = [(x - 2y) - 4/7]^2 = [(x - 2y) - (4/7)][(x - 2y) - (4/7)].
Now we distribute and simplify further:
[(x - 2y) - (4/7)][(x - 2y) - (4/7)]
= (x - 2y - 4/7)(x - 2y - 4/7)
Thus, the factorized form of (x - 2y)^2 - 16/49 is (x - 2y - 4/7)(x - 2y - 4/7).
2. 3/64 - 27y^2:
To factorize this expression, it is important to recognize that it is the difference of squares.
3/64 can be written as (3/8)^2 and 27y^2 as (3y)^2.
So, we have (3/8)^2 - (3y)^2, which can be factored as (3/8 - 3y)(3/8 + 3y).
Thus, the factorized form of 3/64 - 27y^2 is (3/8 - 3y)(3/8 + 3y).