Question 65340
<pre><font size = 5><b>
(4x<sup>3</sup>yz<sup>4</sup>)(2xy<sup>-2</sup>z<sup>-1</sup>)(5x<sup>-2</sup>y<sup>-2</sup>z<sup>3</sup>)

Give the y in the 1st factor and the x in the 
2nd factor an exponent of 1:

(4x<sup>3</sup>y<sup>1</sup>z<sup>4</sup>)(2x<sup>1</sup>y<sup>-2</sup>z<sup>-1</sup>)(5x<sup>-2</sup>y<sup>-2</sup>z<sup>3</sup>)

Multiply the coefficients

4򈮁 = 40

Multiply the x's by adding up the exponents

x<sup>3</sup>x<sup>1</sup>x<sup>-2</sup> = x<sup>3+1-2</sup> = x<sup>2</sup>

Multiply the y's by adding up the exponents

y<sup>1</sup>y<sup>-2</sup>y<sup>-2</sup> = y<sup>1-2-2</sup> = y<sup>-3</sup>

Multiply the z's by adding up the exponents

z<sup>4</sup>z<sup>-1</sup>z<sup>3</sup>  = z<sup>4-1+3</sup> = z<sup>6</sup>

So you have

40x<sup>2</sup>y<sup>-3</sup>z<sup>6</sup>

The probably want you to use only positive
exponents, so move they y<sup>-3</sup> from the
numerator to the denominator as y<sup>3</sup>
<pre><font size = 5><b>
 40x<sup>2</sup>z<sup>6</sup>
棗棗棗棗  
   y<sup>3</sup>

Edwin</pre>