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 $\displaystyle \displaystyle \begin{align*} z + 1 \end{align*}$?

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?

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.

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.

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.

Re: Simplifying polynominals

The last line is incorrect. When you multiply things with the same base, the powers are ADDED, not multiplied.

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.

Re: Simplifying polynominals

Yes that's correct, good job :)

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)