## Denominators with negative degree polynomials

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.

### Denominators with negative degree polynomials

I'm having trouble understanding how to go about simplifying this type of expression. For example, consider 1/((2x^-1)+5); I know that this simplifies to x/(5x+2) in standard form, but why? Is there a rule I'm missing here?
peardog

Posts: 4
Joined: Sat Jul 26, 2014 5:32 pm

### Re: Denominators with negative degree polynomials

peardog wrote:I'm having trouble understanding how to go about simplifying this type of expression. For example, consider 1/((2x^-1)+5); I know that this simplifies to x/(5x+2) in standard form, but why? Is there a rule I'm missing here?

How do you know that this is the correct simplification? What rule(s) did you apply when you attempted the simplification?

nona.m.nona

Posts: 255
Joined: Sun Dec 14, 2008 11:07 pm

### Re: Denominators with negative degree polynomials

I have an answer sheet and I've checked the answer using software. Unfortunately, neither of these showed the steps for simplifying
peardog

Posts: 4
Joined: Sat Jul 26, 2014 5:32 pm

### Re: Denominators with negative degree polynomials

What steps did you try? They show here what negative powers are and how to simplify them. There's more info on simplifying here. And they say here how to do so-called "complex" fractions. So you do the negative power first so the bottom of your "fraction" is (2/x) + 5. Then you simplify the complex fraction.

If you get stuck, please write back showing what you did. Thanks!

maggiemagnet

Posts: 302
Joined: Mon Dec 08, 2008 12:32 am

### Re: Denominators with negative degree polynomials

I posted this as more of a 'proof of concept'. My original problem was ((2x^-1)+5)/((4x^-2)-25). According to Microsoft Mathematics and my answer sheet, the simplified expression is x/(2-5x). Which totally threw me off because I originally tried putting the exponents in their proper place: ((4x^2)+5)/(2x+5) but I ultimately arrived at the wrong answer.

Next I tried simplifying with the odd rule that I apparently know nothing about. For example:
((2x^-1)+5)/((4x^2)-25)
= (x/(2+5x))/((x^2)/(4-(25x^2))
=(x/(5x+2)(*(4-(25x^2))/(x^2)) Then I factored the numerator: (x(5x+2)(5x-2))/((-x^2)(5x+2))
After the simplifying (cancelling out alike terms) I ended up with (5x-2)/(-x) I then multiplied the expression by -1/-1 to get (2-5x)/x
My answer turns out to be identical to the answer given on my answer sheet and in microsoft mathematics, except the numerator and denominator are swapped
(sorry for the nightmare of parentheses)

Edit: The rule I was thinking of was simplifying with like denominators. Thanks for the links
peardog

Posts: 4
Joined: Sat Jul 26, 2014 5:32 pm

### Re: Denominators with negative degree polynomials

Nevermind, my arithmetic was very, very wrong. I solved it. Thanks again for the help
peardog

Posts: 4
Joined: Sat Jul 26, 2014 5:32 pm

### Re: Denominators with negative degree polynomials

peardog wrote:Nevermind, my arithmetic was very, very wrong. I solved it. Thanks again for the help

Thank you for posting the original problem and for showing your work. Congrats on getting it figured out! In case anybody else is interested, the answer looks something like this:

$\frac{2x^{-1} + 5}{4x^{-2} - 25} = \frac{\frac{2}{x} + 5}{\frac{4}{x^2} - 25} = \frac{\frac{2}{x} + \frac{5x}{x}}{\frac{4}{x^2} - \frac{25x^2}{x^2}}$

$= \frac{\left(\frac{2 + 5x}{x}\right)}{\left(\frac{4 - 25x^2}{x^2}\right)} = \left(\frac{2 + 5x}{x}\right)\left(\frac{x^2}{4 - 25x^2}\right) = \frac{x(2 + 5x)}{4 - 25x^2}$

Factor the difference of squares and cancel to get the final answer.

maggiemagnet

Posts: 302
Joined: Mon Dec 08, 2008 12:32 am