SOLUTION: z^2/ z-3 - z/z+3 answer: z3 + 2z^2 + 3z/ (z-3) (z+3) anyone can give me a full break down on why the answer is z3 +2z^2 +3z

Question 193726: z^2/ z-3 - z/z+3

answer: z3 + 2z^2 + 3z/ (z-3) (z+3)
anyone can give me a full break down on why the answer is z3 +2z^2 +3z
Found 2 solutions by jim_thompson5910, RAY100:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Start with the given expression.

Take note that the LCD is . Note: in this case, you simply multiply ALL of the denominators to get the LCD.

Multiply the first fraction by to get the first denominator equal to the LCD

Distribute

Multiply the second fraction by to get the second denominator equal to the LCD

Distribute

Now that the fractions are equal, combine the fractions.

Distribute

Combine like terms.

So where or
Answer by RAY100(1637) (Show Source): You can put this solution on YOUR website!
we believe the problem is:
,
z^2/(z-3) - z/(z+3)
,
LCD is (z-3)(z+3)
,
( z^2(z+3) - z(z-3) ) / ( (z+3) (z-3) )
,
( z^3+3z^2-z^2+3 ) / (z+3)(z-3)
,
( z^3 + 2z^2 +3 ) / ( (z+3)(z-3) )
,
checking,,
let z=0, original =0, answer =0
let z=1, original =-3/4,,,answer = -6/8==-3/4,,,ok
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