anonymous
  • anonymous
(x^2-25/x^2-6x+5)(x^2-1/x^2-2x+1)= x+5/x-2 please explain cuz I got something a lil different
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
Each of these factor, so we have: \[(x-5)(x+5)/(x-1)(x-5) * (x-1)(x+1)/(x-1)(x-1)\] the (x-5)'s cancel out on the first one, and the (x-1)'s cancel out on the 2nd one
campbell_st
  • campbell_st
\[((x - 5)(x +5))/((x - 5)(x+1)) = (x + 5)/(x + 1)\] \[((x + 1)(x-1))/(x - 1)(x -1)) = (x+1)/(x-1)\] (x + 10 will cancel leaving (x+5)/(x-1)
anonymous
  • anonymous
\[(x^2-25)(x^2-1)=(x+5)(x-5)(x-1)(x+1)\] \[(x^2-6x+5)(x^2-2x+1)=(x-5)(x-1)(x-1)(x-1)\]

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anonymous
  • anonymous
so we have \[(x+5)/(x-1) * (x+1)/(x-1)\]
anonymous
  • anonymous
@campbell you messed up on your factoring
anonymous
  • anonymous
Got that does it some how reduce further to get x+5/x-2?
anonymous
  • anonymous
I wouldnt think so but I could be wrong
anonymous
  • anonymous
That's what I got too but then the book says the x+5/x-2 answer
anonymous
  • anonymous
I'm thinking the book got it wrong.
campbell_st
  • campbell_st
x^2 - 2x + 1 is a perfect square (x-1)^2 or (x -1)(x-1)
anonymous
  • anonymous
Think soo too cuz I know I did the process right and if you came up with the same answer then ya
anonymous
  • anonymous
@campbell I meant on the first denominator you messed up
anonymous
  • anonymous
book can be wrong again but wanna make sure.... x^2-x+11 (prime right)
anonymous
  • anonymous
Yes, thats prime

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