# Simplifying polynominals

• Jul 5th 2013, 08:35 PM
Welkin
Simplifying polynominals
Hi all, I've gone through the 'read before' threads so I hope this is in the right section and that I've typed it out ok, I'm still working with pen and paper here (Happy)

Basically I've just started to get back into mathematics and I'm doing a course externally. I've had a try at simplifying the problem below but all the rules are a very new to me and so I have no idea if I'm on the right track or not. Any feedback at all would be a huge help, Thanks.

Simplify:

(z + 1)^2 (3xy^2)^3 / (9y^2)^1/2 x^-3 (z + 1) ^-1

= (z + 1)^2 (9x^3y^6) / 3y x^-3 (z + 1)^-1

= (z + 1) ^2 9x^3y^6 1/3y^-1 x^3 (z + 1)

= (z + 1)^3 9x^9y^6 1/3y^-1

= (z + 1)^3 3x^9y^-6

I've typed it out as I've done the steps because I'd almost guarantee the final answer is wrong so like I said any feedback on what I've done right or wrong would be a huge help.
• Jul 5th 2013, 08:42 PM
Prove It
Re: Simplifying polynominals
For starters, \displaystyle \displaystyle \begin{align*} 3^3 = 27 \end{align*}, not 9.

Also how did you magically turn \displaystyle \displaystyle \begin{align*} y \end{align*} into \displaystyle \displaystyle \begin{align*} y^{-1} \end{align*}, \displaystyle \displaystyle \begin{align*} x^{-3} \end{align*} into \displaystyle \displaystyle \begin{align*} x^3 \end{align*} and \displaystyle \displaystyle \begin{align*} (z + 1)^{-1} \end{align*} into \displaystyle \displaystyle \begin{align*} z + 1 \end{align*}?
• Jul 5th 2013, 08:49 PM
Welkin
Re: Simplifying polynominals
Quote:

Originally Posted by Prove It
For starters, \displaystyle \displaystyle \begin{align*} 3^3 = 27 \end{align*}, not 9.

Also how did you magically turn \displaystyle \displaystyle \begin{align*} y \end{align*} into \displaystyle \displaystyle \begin{align*} y^{-1} \end{align*}, \displaystyle \displaystyle \begin{align*} x^{-3} \end{align*} into \displaystyle \displaystyle \begin{align*} x^3 \end{align*} and \displaystyle \displaystyle \begin{align*} (z + 1)^{-1} \end{align*} into \displaystyle \displaystyle \begin{align*} z + 1 \end{align*}?

Thanks, cant believe I missed the first part considering how many times I've looked at it.

For the rest I thought that if something was to the power of -1 below the dividing line then it can be moved to the top (numerator section?) and would become a positive to the same power. I was trying to do the same for 3y, moving it above the line as 1/3y^-1?
• Jul 5th 2013, 08:57 PM
Welkin
Re: Simplifying polynominals
So would:

(z + 1)^2 (27x^3y^6)
-----------------------
3y x^-3 (z + 1)^-1

be a correct first step? (sorry I couldn't find any suggestions about the best way to type out a problem). If it is ok I suppose I'm stuck there on how to simplify it further.
• Jul 5th 2013, 09:43 PM
Prove It
Re: Simplifying polynominals
Yes it would be. And to typeset mathematics, we use LaTeX. There is an inbuilt compiler on this site and a subforum dedicated to learning how to use it and gaining assistance with it.

As for negative powers, \displaystyle \displaystyle \begin{align*} a^{-m} = \frac{1}{a^m} \end{align*}, so whenever you see negative powers, you can assume there is division. With some manipulation, this would be equivalent to moving everything with a negative power from the top to the bottom with a positive power, or vice versa.
• Jul 5th 2013, 10:23 PM
Welkin
Re: Simplifying polynominals
Ok so:

Original question:

$\displaystyle \frac{(z+1)^2 (3xy^2)^3}{(9y^2)^{\frac{1}{2}} x^{-3} (z+1)^{-1}}$

= $\displaystyle \frac{(z + 1)^2 (27x^3y^6)}{3y x^{-3} (z + 1)^{-1}}$

So would this also be correct:

= $\displaystyle (z + 1)^2 27x^3y^6 \frac{1}{3}y^{-1}x^3 (z+1)$

= $\displaystyle (z+1)^3 9x^9y^{-6}$

and if it is correct, is it correct as far as simplifying?

Thanks for the help so far.
• Jul 5th 2013, 11:25 PM
Prove It
Re: Simplifying polynominals
The last line is incorrect. When you multiply things with the same base, the powers are ADDED, not multiplied.
• Jul 5th 2013, 11:52 PM
Welkin
Re: Simplifying polynominals
So many silly mistakes, I did it that way the first few dozen times, honest

$\displaystyle (z+1)^3 9x^6y^{-5}$

This is going to take some work to get familiar with... though no more than before I started I suppose.
• Jul 5th 2013, 11:59 PM
Prove It
Re: Simplifying polynominals
Yes that's correct, good job :)
• Jul 7th 2013, 08:00 PM
Welkin
Re: Simplifying polynominals
Quote:

Originally Posted by Welkin
So many silly mistakes, I did it that way the first few dozen times, honest

$\displaystyle (z+1)^3 9x^6y^{-5}$

This is going to take some work to get familiar with... though no more than before I started I suppose.

$\displaystyle (z+1)^3 9x^6y^{5}$ ..... I noticed later that I had left the y power negative and it'd bug me if I didn't come back and change it (Happy)