math help?

Question

algebraic fractions
(2y-80)^2/(2y-8)^3
please show me what to do?

Answer 1

(a - b)^2 = a^2 - 2ab + b^2

Using that formula, you just simplify.
Let's start with the numerator.

(2y - 80)^2
(2y)^2 - 2(2y)(80) + (80)^2
4y^2 - 320y + 6400

Do the same thing with the denominator.
(2y - 8)^3 = (2y - 8)(2y - 8)(2y - 8) = (2y -8)(2y - 8)^2
(2y -8) [(2y)^2 - 2(2y)(8) + (8)^2]
(2y - 8)(4y^2 - 32y + 64)

Now just foil that out.
8y^3 - 64y^2 + 128y - 32y^2 + 256y - 512
8y^3 - 96y^2 + 384y - 512

Put the numerator over the denominator.

(4y^2 - 320y + 6400) / (8y^3 - 96y^2 + 384y - 512)

This can be simplified, though, if you factor out a 4 on both the numerator and denominator, then cancel them out.

The final answer is

(y^2 - 80y + 1600) / (2y^3 - 24y^2 + 96y - 128)

Hope that helps!

Answer 2

are the ^ multiplications signs?
if so the here we go
(2y-80)*2/(2y-8)*3 distribute
4y-160/6y-24
that is the simplified answer...i cant do anymore without it being a equation.

Answer 3

monkey.. is right about what you have there. But i understand what raj was saying, if the 0 is your typo, it would be a lot easier.
(2y-8)^2-3
(2y-8)^-1
1/(2y-8)

Answer 4

(2y-80)² over (2y-8)³
(2y - 80)(2y - 80) : (2y - 8)(2y - 8) (2y - 8)
(4y² - 320y + 6400) : 8y² - 512
4(y² - 80y + 1600) : 8(y² - 64) =
Simplyfing 4 and 8
(y² - 80y + 1600) : 2(y² - 64)
<>>

Answer 5

please check wherther you have posted the question correctly?