HELP!!! I dont remember how to simplify!!!?

Question

I don't remember how to simplify these can anyone help me...i want the answer but i WANT YOU TO EXPLAIN how you got the answer. I want to know how to simplify. Thank you sooo much!

x ^-3

--------

x ^-5

2x^3 y^2

-------- * ------

y ^-5 3x^7

x^2-4x-5

-------------

x^2+2x+1

x-4

-----

4-x

(x-1)^3

x^-1/3 x^-3/5

(For the last one I'm not sure if the negative sign is for the whole fraction or for just the denominator...on my paper it just comes right after the x...but the negative and the fraction are exponents)

Hopefully you understand lol...b/c i dont...

Answer 1

You have to remember the rules of exponents:

(x^a)(x^b) = x^(a + b)

(x^a)/(x^b) = x^(a - b)

x^0 = 1

x^(-a) = 1/(x^a)

x^(a/b) = the b-th root of x^a:  for instance, x^(2/3) = the cube root of x squared.

(x^a)^b = x^(ab)

You can only simplify factors, not terms.  So, for instance,

x/(xy) = 1/y as the x's cancel out

x/(x + y) is as simple as it gets.

So:

(x^(-3))/(x^(-5)) = x^(-3 - (-5)) = x^2

[(2x^3)/(y^(-5))] * [(y^2)/(3x^7)]

Multiplying fractions:  (a/b)*(c/d) = (ab)/(cd)

So,

(2x^3 y^2)/(3 x^7 y^(-5)) = (2/3)(x^3/x^7)(y^(-5)/y^2)) =

(2/3) (x^(-4))(y^(-7)) = 2/(3 x^4 y^7)

Pulling the negative exponents into the denominator makes them positive.  Pushing negative exponents in the denominator into the numerator makes them positive.  Again, this only works with factors, not terms.

(x^2 - 4x - 5)/(x^2 + 2x + 1)

All we have here are terms, so we're going to have to factor the quadratics before we can go further.

(x^2 - 4x - 5) = (x - 5)(x + 1)

(x^2 + 2x + 1) = (x + 1)(x + 1)

When we put the numerator over the denominator, we have factors of (x + 1) in both.  They cancel (because (x + 1)/(x + 1) = 1, and 1 changes nothing), so we end up with (x - 5)/(x + 1)

(x - 4)/(4 - x)  Again, we have terms, not factors, and both are linear so we can't factor them.  However, notice that (x - 4) = -(4 - x).  So we have

-(4 - x)/(4 - x) = -1

(x - 1)^3

If this is (x^(-1))^3, then it's x^(-3) = 1/(x^3)

If it's really (x - 1)^3, then it's just a matter of multiplying it out.  This can be done quite quickly if you know Pascal's Triangle (google it if you don't):  (x - 1)^3 = x^3 - 3x^2 + 3x - 1

Otherwise, just take it two at a time:  find (x - 1)(x - 1) and multiply the result by (x - 1)

x^(-1/3)x(-3/5) = x^(-1/3 + -3/5) = x^(-14/15) = 1/(x^(14/15))

Learn those rules.  Save you they can.  (There are easy ways to remember or derive them, but it's too complicated to go into here.  Ask your teacher why those rules are what they are.)

Edit:  Just one other thing.  Normally, I wouldn't have answered this question.  There were too many questions all rolled into one, which as you can see takes some time.  I decided to go ahead because of your TC badge.  I figure if you're willing to spend the time answering other people's questions, you deserve a little more consideration.  Think of it as a benefit of the TC thing.  God knows it's the only one you'll get.  :)