Question 134835
[(2x^2 - 3x - 9)/ (2x^2 - 18)] *[(x^2 + x - 6)/(2x^2 - x -6)]

Factor where you can:

= [(2x+3)(x-3)/2(x-3)(x+3)] * [(x+3)(x-2)/(2x+3)(x-2)]

Cancel factors common to num. and denom.: (2x+3),(x+3).(x-3),(x-2)

=  [1/2] * [1/1]

= 1/2

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Cheers,
Stan H.



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