math help!! ASAP?

Question

x^2 + 5x
------------ <- division line
x^2 - 25

how do u do that...i did (a+b) * (a-b) at the bottom becuz it's perfect squares but after that im stuck.

2y^3 - 12y^2 + 2y
----------------------
y^2 - 6y + 1

im completely stuck there. these questions are on a sheet that asks us to use the formula ax^2 + bx +c

PS: ^ <- this sing means to the power of....y^2 means Y squared

THANKS A LOT EVERY1... but there's one more problem i dont get....it's this kind.

X^2 + 4X
-----------
X^2 - 9X

im thinking of getting rid of the X on the 2nd # so it makes perfect squares and u can do (a+b) * (a-b) but that would mean that the X^2 will end up being just X so it wont work......can u just factor X so it ends up being :

x (x+4)
---------
x (x - 9)

Answer 1

x^2+5x --> x(x+5) --> x / (x-5)
--------- --> -----------
x^2-25 --> (x+5)(x-5)


2y^3 - 12y^2 + 2y --> 2y(y^2 - 6y + 1)
---------------------- --> ------------------- --> 2y
y^2 - 6y + 1 --> (y^2 - 6y +1)

Answer 2

you did great on the bottom: (x + 5)(x - 5).
Now look at the numerator; those two terms have an 'x' in common. Factor it out and you get x(x + 5)

Now, the (x+5) will cancel with the one on the bottom, giving you x/(x-5).

The second problem has a 2y common factor in the numerator, and both the top and the bottom can be factored. Good luck.

Answer 3

The top one: remove the x so its: x(x+5). The bottom is (x+5)(x-5). You can remove the x+5 from the top and bottom because they cancel, right? so the answer is x/x-5.

The bottom one: You remove 2y out so its y^2-6y+1. Now, divide the common terms so your answer is: 2y. What grade are you in? It looks like grade 10 stuff. The stuff I taught last year! Have fun.

Answer 4

Here's no.1. Number 2 works the same way.
Take a common factor out of the numerator and
factor the denominator to get
x(x+5)/(x+5)(x-5)
Now cancel the common factor of x+5 to get
x/(x-5). This is valid for all x except x = 5 or -5.
For the second one, don't try to factor the
bottom. Just take a common factor out of the top
and see what happens.
Hope this helps a bit.

Answer 5

You need to factor on this first one.

x(x+5) / (x+5)(x-5)

then you can cancel the (x+5) in the numerator and the denomonator.

this will leave you x/(x-5)

The next one .... again factor

2y(y^2-6y+1) / (y^2-6y+1)

then the y^2-6y+1 cancels leaving you 2y

Answer 6

1) (x^2+5x)/(x^2-25) = (x(x+5)) / ((x+5)(x-5)) = x/(x-5)

2) (2y^3-12y^2+2y)/(y^2-6y+1)=(2y(y^2-6y+1)) / (y^2-6y+1) = 2y

hope that clear enough.. lol..too bad this yahoo answer not provided with mathemtical operational formula >.<

Answer 7

the first one just factor the denominator
x(x+5)/((x-5)(x+5) cancel out u get
=> x/(x-5)

second one

2y(y^2 -6y+1)/(y^2 - 6y + 1
)
cancl out you will get 2y

Answer 8

Ah! My exact area of expertise! But, I can't tell you the answer because math is one of those things you need to learn and understand on your own. If your book can't help you, ask your teacher. They'll explain it to you rationally and make sure you understand it without giving you the answer.

Answer 9

oopps... gotto do it this way...
x (x+5)
------------- =
(x+5)(x-5)

x / (x-5)

Answer 10

i donnt know...but it scares me =(