Question 805633
x^2-12x+27 over x^2-15x+56 * x^2-19x+84 over x^2-20x+99
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Factor where you can------
[(x-3)(x-9)]/[(x-7)(x-8)] * [(x-7)(x-12)]/[(x-9)(x-11)]
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Cancel all factors common to a numerator and a denominator, such as
(x-9) and (x-7), to get::
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= [(x-3)]/[(x-8)] * [(x-12)]/[(x-11)]
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= [[(x-3)(x-12)]/[(x-8)(x-11)]
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= (x^2-15x+36)/(x^2-19x+88)
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Cheers,
Stan H.
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Answers are:: 
A. (x+3)(x+12) over (x+8)(x+11)
B. (x^2-12x+27)(x^2-19x+84) over (x^2-15x+56)(x^2-20x+99)
C. (x-3) over (x-11)
D. (x-3)(x-12) over (x-8)(x-11)