another polynomial

**tigersblood** · July 15th 2007, 07:42 AM

Thank you all for your help today! Here's another that's been stumping me:

(2X¯¹Y²)(4X²Y¯³)¯²
(12X¯²Y¯²)¯¹

**red_dog** · July 15th 2007, 08:16 AM

$\frac{(2x^{-1}y^2)(4x^2y^{-3})^{-2}}{(12x^{-2}y^{-2})^{-1}}=\frac{2x^{-1}y^2(2^2)^{-2}(x^2)^{-2}(y^{-3})^{-2}}{(2^2\cdot 3)^{-1}(x^{-2})^{-1}(y^{-2})^{-1}}=\frac{2x^{-1}y^22^{-4}x^{-4}y^6}{2^{-2}3^{-1}x^2y^2}=$
$=\frac{2^{-3}x^{-5}y^8}{2^{-2}3^{-1}x^2y^2}=\frac{3}{2}x^{-7}y^6$

**tigersblood** · July 15th 2007, 08:26 AM

The answer in my booklet is

3X
2Y^7

**tigersblood** · July 15th 2007, 08:28 AM

Sorry, I stand corrected:

3X
2Y^6

**Soroban** · July 15th 2007, 08:38 AM

Hello, tigersblood!

The answer in my booklet is: . $\frac{3x}{2y^7}$

Either you gave us the wrong problem . . . or your booklet is dreadfully wrong.

**tigersblood** · July 15th 2007, 08:43 AM

I just checked and rechecked the problem. I copied it correctly, and the answer in the back is
3x
2y^6

Aargh.

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