Divide (x^4-y^4) by (x-y).
Need to show work to help me understand.
Also need to answer without parentheses.
Thanks for any and all help with this one!!
(x^4-y^4)
= (x^2 - y^2)(x^2 + y^2)
= (x+y)(x-y)(x^2 + y^2)
so (x^4-y^4)÷(x-y)
= (x+y)(x-y)(x^2 + y^2)÷(x-y)
= (x+y)(x^2 + y^2)
expand this if you don't want the brackets.
Hey, thanks for your help.
welcome
To divide (x^4-y^4) by (x-y), you can use the concept of factoring the difference of squares. The difference of squares formula states that a^2 - b^2 can be factored as (a + b)(a - b).
In this case, we have (x^4 - y^4) in the numerator and (x - y) as the denominator. To simplify, we need to factor the numerator as a difference of squares.
The numerator (x^4 - y^4) can be written as (x^2)^2 - (y^2)^2. Applying the difference of squares formula, we get (x^2 + y^2)(x^2 - y^2).
Now, we can further factor (x^2 - y^2) as (x + y)(x - y).
Therefore, the fully factored expression becomes:
(x^4 - y^4) = (x^2 + y^2)(x + y)(x - y)
To divide by (x - y), we can cancel the common factor of (x - y) in the numerator and the denominator:
(x^4 - y^4) / (x - y) = (x^2 + y^2)(x + y)(x - y) / (x - y)
Canceling the common factor (x - y), we are left with the simplified expression:
(x^4 - y^4) / (x - y) = (x^2 + y^2)(x + y)
So, the answer to the division problem (x^4 - y^4) / (x - y) without parentheses is (x^2 + y^2)(x + y).