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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