please help. Divide ((x^2+2x-15)/4x^2)) by ((x^2-25)/2x-10))?

Question

also, ((x^2-9y^2)/(7x^2-21y)) by (x^2+3xy)

Thank you. I'm so stuck on these 2....

Answer 1

Divide ((x²+2x-15)/4x²)) by ((x²-25)/2x-10))

Let's start by simplifying the second term, ((x²-25)/2x-10)).

((x²-25)/2x-10))
[(x + 5)(x - 5)] / [2(x - 5)]
Cancel out the (x - 5)
(x + 5) / 2

((x²+2x-15)/4x²))
Simplifying:
[(x + 5)(x - 3)] / (4x²)

[(x + 5)(x - 3)] / (4x²) divided by [(x + 5) / 2] =

[(x + 5)(x - 3)] / (4x²) times [2 / (x + 5)] =

[(x + 5)(x - 3)][2] / (4x²)(x + 5)

Cancel out the (x + 5)

[(x - 3)][2] / (4x²) =



Final answer:
(x - 3) / (2x²)

and x cannot equal ±5, since we cancelled out both (x - 5) and (x + 5) from the denominator, and also, x cannot equal zero.


Second problem,

((x²-9y²)/(7x²-21y)) by (x²+3xy)

Factoring:

[(x + 3y)(x - 3y)] / [7(x²-3y)] times [1 / (x²+3xy)]

[(x + 3y)(x - 3y)]/ [7(x²-3y)x(x+3y)]

Cancel out (x + 3y) gives you

[(x - 3y)]/ [7x(x²-3y)]

I guess you could also further factor this, into:

[(x - 3y)]/ [7x(x + √[3y])(x - √[3y])]

Answer 2

One to a customer with all this noise. Division by the second expression is acheived by multiplication by its reciprocal. So you would have
(x^2+2x-15) * (2x-10) DIVIDED BY
(4x^2) * ( x^2 -25 )

I suspect that once you factor the quadratic in the numerator, and take the difference of squares of the the one in the denominator, alot of stuff will divide out to a much more simple form.