How do you divide #{(n^2-n-12)/(2n^2-15n+18)}/{(3n^2-12n)/(2n^3-9n^2)}#?

1 Answer
Shwetank Mauria
Apr 3, 2016

#(n(n+3)(2n-9))/(3(2n-3)(n-6))#

Explanation:

#((n^2-n-12)/(2n^2-15n+18))/((3n^2-12n)/(2n^3-9n^2))# is equivalent to #(n^2-n-12)/(2n^2-15n+18)xx(2n^3-9n^2)/(3n^2-12n)#, as a division by a fraction is equivalent to multiplication by its multiplicative inverse.

For solving the above we need to factorize each polynomial.

#n^2-n-12=(n^2-4n+3n-12=n(n-4)+3(n-4)=(n+3)(n-4)#

#2n^2-15n+18=2n^2-12n-3n+18=2n(n-6)-3(n-6)=(2n-3)(n-6)#

#3n^2-12n=3n(n-4)# and #2n^3-9n^2=n^2(2n-9)#

Hence, #(n^2-n-12)/(2n^2-15n+18)xx(2n^3-9n^2)/(3n^2-12n)# is equivalent to

#((n+3)cancel((n-4)))/((2n-3)(n-6))xx(n(cancel(n^2))(2n-9))/(3cancelncancel((n-4)))#

= #(n(n+3)(2n-9))/(3(2n-3)(n-6))#

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