SOLUTION: 1-y^2/y^2+4y+3 and 5x^2-125/2x^2-9x-5

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: 1-y^2/y^2+4y+3 and 5x^2-125/2x^2-9x-5      Log On


   



Question 1001737: 1-y^2/y^2+4y+3
and
5x^2-125/2x^2-9x-5

Answer by ikleyn(52805) About Me  (Show Source):
You can put this solution on YOUR website!
.
There is no question in your submitting.
Also there is no necessary parentheses in it.
It is incomplete.
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In opposite, if some of these circumstances are not in the place,
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Then think what was your mistake and how to fix it.  You can resubmit your request.

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Comment from student:  The question is to simplify those two problems.
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OK,  let's do it.

1.  %281-y%5E2%29%2F%28y%5E2%2B4y%2B3%29.

    The polynomial in the numerator can be factored  1-y%5E2 = (1+y)*(1-y).

    The polynomial in the denominator can be factored  y%5E2%2B4y%2B3 = (y+1)*(y+3).

    Thus  %281-y%5E2%29%2F%28y%5E2%2B4y%2B3%29 = %28%281%2By%29%2A%281-y%29%29%2F%28%28y%2B1%29%2A%28y%2B3%29%29 =       (reduce the numerator and the denominator by the factor (1+y) )   ----->

    = %28cross%28%281%2By%29%29%2A%281-y%29%29%2F%28cross%28%28y%2B1%29%29%2A%28y%2B3%29%29 = %281-y%29%2F%28y%2B3%29.


2.  %285x%5E2-125%29%2F%282x%5E2-9x-5%29.

    The numerator can be factored  5x%5E2-125 = 5*(x+5)*(x-5).

    The denominator can be factored  2x%5E2-9x-5 = (x-5)*(2x+1).     (You can get this factoring by finding the roots of the quadratic polynomial).

    Thus  %285x%5E2-125%29%2F%282x%5E2-9x-5%29 = %285%2A%28x-5%29%2A%28x%2B5%29%29%2F%28%28x-5%29%2A%282x%2B1%29%29 =       (reduce the numerator and the denominator by the factor (x-5) )   ----->

    = %285%2Across%28%28x-5%29%29%2A%28x%2B5%29%29%2F%28cross%28%28x-5%29%29%2A%282x%2B1%29%29 = %285%2A%28x%2B5%29%29%2F%282x%2B1%29.