Fractional Algebra... Exponents?

Hi,

2x^5y^2 - 4xy^2
----------------------- =
. . . . 2xy^2

2x^5y^2 . 4xy^2
----------- - --------- =
2xy^2 . . .2xy^2

Divide coefficients. Subtract exponents on each variable.

2x^5y^2 . 4xy^2
----------- - --------- = x^4 - 2 <==ANSWER
2xy^2 . . .2xy^2

I hope that helps!! :-)

= (2x^5y^2 - 4xy^2)/(2xy^2)
= x^4 - 2

Answer: x^4 - 2

Checking back to 2x^5y^2:
= (x^4 - 2)(2xy^2)
= 2x^5y^2 - 4xy^2

Yes, when you multiply the same base times itself you add the exponents. Think about it, it makes sense. 5^2 * 5^3 = ? 5*5 * 5*5 * 5 = 5^5 5^(2+3) = 5^5 A fractional exponent is: The numerator is the base to that power and the denominator is the base to that root. 5 ^(2/3) = cube root of 5^2 Your example (X^6 * Y^3) ^1/3 = cube root of x^6 = x^2 cube root of Y^3 = Y (x^2*Y) Oh yeah bet they never told you Any number to 0 power = 1 Any number to 1 power = the number That is why (x^6)^1/3 following the rule x^6 is raised to the first power which is just x^6 them we take the cube root of that.

( 2x ) ( y² ) ( x^4 - 2 )
-----------------------------
( 2x ) ( y² )

x^4 - 2

( x² - √2 ) ( x² + √2 )

(2x^5y^2 - 4xy^2)/2xy^2
= 2xy^2(x^4 - 2)/2xy^2
= x^4 - 2

(2x^5y^2 - 4xy^2) / 2xy^2
x^4 - 2 Simplify by using the law of exponents.

You subtract the exponents and then divide the coefficiants.

My answer is x^4 - 2 WHY?

2x^5y^2 / 2xy^2 divide the number coefficient left is 1, divide the letter coefficient, cancel x left is x^4 and lastly cancel y
so 2x^5y^2 / 2xy^2 = 1x^4 or simplify = x^4

-4xy^2 / 2xy^2 divide the number coefficient again left is -2, divide the letter coefficient, cancel the xy^2
so -4xy^2 / 2xy^2 = -2

therefore:

2x^5y^2 - 4xy^2 / 2xy^2 =x^4 - 2

(2x⁵y² - 4xy²)/(2xy²) = (2xy²(x⁴ - 2))/(2xy²) = x⁴ - 2

This is the same as saying [(2x^5y^2)/2xy^2] - (4xy^2/2xy^2) I'm assuming, unless the problem is 2x^5y^2-(4xy^2 / 2xy^2). Using parentheses is a great help, as you can see the way you posted it is ambiguous. I'm going with the first one. Just divide/reduce each term:

2x^5y^2/2xy^2

2x^5/2x leaves x^4
y^2/y^2 leaves 1, so the term is x^4

4xy^2/2xy^2
4x/2x=2
y^2/y^2=1, so you're left with 2. The equation is now:

x^4-2. That is your answer.

wpf.

= (2x^5y² - 4xy²) / (2xy²)

= [2xy² (x^4 - 2)] / (2xy² )

= (x^4 - 2) ANSWER.