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anonymous

  • 4 years ago

(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

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  1. anonymous
    • 4 years ago
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    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

  2. campbell_st
    • 4 years ago
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    \[((x - 5)(x +5))/((x - 5)(x+1)) = (x + 5)/(x + 1)\] \[((x + 1)(x-1))/(x - 1)(x -1)) = (x+1)/(x-1)\] (x + 10 will cancel leaving (x+5)/(x-1)

  3. anonymous
    • 4 years ago
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    \[(x^2-25)(x^2-1)=(x+5)(x-5)(x-1)(x+1)\] \[(x^2-6x+5)(x^2-2x+1)=(x-5)(x-1)(x-1)(x-1)\]

  4. anonymous
    • 4 years ago
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    so we have \[(x+5)/(x-1) * (x+1)/(x-1)\]

  5. anonymous
    • 4 years ago
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    @campbell you messed up on your factoring

  6. anonymous
    • 4 years ago
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    Got that does it some how reduce further to get x+5/x-2?

  7. anonymous
    • 4 years ago
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    I wouldnt think so but I could be wrong

  8. anonymous
    • 4 years ago
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    That's what I got too but then the book says the x+5/x-2 answer

  9. anonymous
    • 4 years ago
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    I'm thinking the book got it wrong.

  10. campbell_st
    • 4 years ago
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    x^2 - 2x + 1 is a perfect square (x-1)^2 or (x -1)(x-1)

  11. anonymous
    • 4 years ago
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    Think soo too cuz I know I did the process right and if you came up with the same answer then ya

  12. anonymous
    • 4 years ago
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    @campbell I meant on the first denominator you messed up

  13. anonymous
    • 4 years ago
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    book can be wrong again but wanna make sure.... x^2-x+11 (prime right)

  14. anonymous
    • 4 years ago
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    Yes, thats prime

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