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8th Grade Algebra

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(b^2-10b+21)/(b^2-11b+28)

I need help trying to figure out how to factor each quantity or parentheses.

and then simplifying it.

  • 8th Grade Algebra -

    For (b^2-10b+21), start with listing the factors that multiply to give 21.
    1,21
    3,7
    7,3
    21,1
    As 21 is positive and -10b is negative, you will find that the factors are in the following form:
    (b-x)(b-y)
    where x and y are chosen from the pair of factors. If you multiply out the factors, you will get
    (b-x)(b-y)=b^2-(x+y)b+xy
    Thus, you will choose from the list of factors a pair which will give
    xy=21 and x+y=10.
    Try the same for the denominator. It has more factors, so the list of factors will look like:
    1,28
    2,14
    4,7
    7,4
    14,2
    28,1
    The method to follow is exactly the same as the numerator.
    Tell us what you found.

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