Question

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

Answer 1

`(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))`