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Bananas1234

  • one year ago

Simplify and identify the domain. -5/5x + 15 times 2(x + 2)/x^2 - 4

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  1. rajat97
    • one year ago
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    to identify the domain, we need to see at which points, the function is undefined so the function is undefined at three points. those points are -3,2 and -2 the function is defined and exists at all real numbers except these three. thus the domain is all real numbers except -3,-2,2

  2. rajat97
    • one year ago
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    hello Bananas1234 is it right??

  3. UsukiDoll
    • one year ago
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    I think everyone needs a mandatory course in latex or something. I can't read this.

  4. UsukiDoll
    • one year ago
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    rajat, you're right for -2,2, and -3 will make the denominator 0. assuming that this is the right equation.

  5. Bananas1234
    • one year ago
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    idk what latex is. and i agree with the -2, 2 and -3 but what about the simplify part?

  6. Bananas1234
    • one year ago
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    @UsukiDoll

  7. anonymous
    • one year ago
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    factor both denominators, then cancel common terms

  8. rajat97
    • one year ago
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    look at the denominators

  9. rajat97
    • one year ago
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    the numbers at which the denominators become zero are the numbers that will be excluded from the domain

  10. Bananas1234
    • one year ago
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    |dw:1433046092010:dw|

  11. UsukiDoll
    • one year ago
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    -5/5x + 15 times 2(x + 2)/x^2 - 4 \[\frac{-5}{5x+15} \times \frac{2(x+2)}{x^2-4}\]

  12. UsukiDoll
    • one year ago
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    if we can pull that 5 out on the first denominator and then make sure that \[x^2-4 = (x+2)(x-2) \] omg latex come on. x^2-4 = (x+2)(x-2) cancel out some of the terms... you should be able to cancel out 2 of them

  13. UsukiDoll
    • one year ago
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    sorry my end is bugged... need to refresh

  14. UsukiDoll
    • one year ago
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    \[\frac{-5}{5(x+3)} \times \frac{2(x+2)}{(x+2)(x-2)}\] do you see anything that needs to be canceled out?

  15. UsukiDoll
    • one year ago
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    the common terms are 5 and x+2... since I can see them on the numerator and denominator we can cancel it out.

  16. UsukiDoll
    • one year ago
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    \[\frac{-1}{(x+3)} \times \frac{2}{(x-2)}\] and since we're not dealing with addition, we can just multiply this through. \[\frac{-2}{(x+3)(x-2)}\]

  17. Bananas1234
    • one year ago
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    Thanks

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