Question 78332
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Factor the numerator


*[invoke quadratic_factoring 3, 9, -30]



Now factor the denominator



*[invoke quadratic_factoring 3, 0, -75]




So the expression {{{(3x^2+9x-30)/(3x^2-75)}}} factors to


{{{((3x-6)(x+5))/((3x+15)(x-5))}}}


{{{((3x-6)(x+5))/(3(x+5)(x-5))}}} Factor out a 3 in the denominator


{{{((3x-6)cross((x+5)))/(3cross((x+5))(x-5))}}} Notice these terms cancel


{{{(3x-6)/(3(x-5))}}}


{{{3(x-2)/(3(x-5))}}} Factor out a 3 in the numerator


{{{cross(3)(x-2)/(cross(3)(x-5))}}} Notice the 3's cancel


{{{(x-2)/(x-5)}}} This is the simplified answer