The difference of two squares factoring pattern states that a difference of two squares can be factored as follows:
So, whenever you recognize the two terms of a subtraction to be two squares, you can factor it as the sum of the roots multiplied by the difference of the roots.
In this case, the squares are obvious: 
is the square of 
, and 
is the square of 
So, we can factor the expression as
![(x+2)^2 - (y+2)^2 = [(x+2)+(y+2)] - [(x+2)+(y+2)]](https://tex.z-dn.net/?f=%20%28x%2B2%29%5E2%20-%20%28y%2B2%29%5E2%20%3D%20%5B%28x%2B2%29%2B%28y%2B2%29%5D%20-%20%5B%28x%2B2%29%2B%28y%2B2%29%5D%20)
(the round parenthesis aren't necessary, I used them only to make clear the two terms)
We can simplify the expression summing like terms:
![(x+2)^2 - (y+2)^2 = [(x+2)+(y+2)][(x+2)-(y+2)] = (x+y+4)(x-y)](https://tex.z-dn.net/?f=%28x%2B2%29%5E2%20-%20%28y%2B2%29%5E2%20%3D%20%5B%28x%2B2%29%2B%28y%2B2%29%5D%5B%28x%2B2%29-%28y%2B2%29%5D%20%3D%20%28x%2By%2B4%29%28x-y%29%20)
(x-2)squared +y squared=20
Answer:
false
Step-by-step explanation:
Oh that question
Parallel sides
Opposite sides are congruent
And diagonals bisect each other
Answer:
An expression for Francisco's age in terms of a is (2a+3)
An expression for Harrison's age in terms of a is (3a-2)
An expression for the sum of all three ages, in terms of a is (6a+1)
Step-by-step explanation:
Let
a -----> Jenna's age
b -----> Francisco's age
c -----> Harrison's age
we know that
Francisco's age is two times Jenna's age plus three years
b=2a+3 -----> equation A
Harrison's age is three times Jenna's age minus two years
c=3a-2 ----> equation B
The sum of all three ages, in terms of a is
(a+b+c) -----> equation C
substitute equation A and equation B in equation C
a+(2a+3)+(3a-2)
Combine like terms
(a+2a+3a)+(3-2)
6a+1
therefore
An expression for Francisco's age in terms of a is (2a+3)
An expression for Harrison's age in terms of a is (3a-2)
An expression for the sum of all three ages, in terms of a is (6a+1)
