How do you divide #(4k^2-29k-24)/(k^2-13k+40)# by #(4k^2+15k+9)/(5k^3-25k^2-3k+15)#?

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Answer 1 · 2016-12-04T20:10:40

#((5k^2-3))/((k+3))#

With algebraic fractions, the first approach is to factorise where possible:

#(4k^2-29k-24)/(k^2-13k+40) div(4k^2+15k+9)/(5k^3-25k^2-3k+15)#

#=((4k+3)(k-8))/((k-8)(k-5)) div ((4k+3)(k+3))/(5k^2(k-5) -3(k-5))#

To divide by a fraction, multiply by the reciprocal.

#=((4k+3)(k-8))/((k-8)(k-5)) xx ((k-5)(5k^2 -3))/((4k+3)(k+3))#

Now cancel like factors

#=(cancel((4k+3))cancel((k-8)))/(cancel((k-8))cancel((k-5))) xx (cancel((k-5))(5k^2 -3))/(cancel((4k+3))(k+3))#

#=((5k^2-3))/((k+3))#