How do you divide {(n^2-n-12)/(2n^2-15n+18)}/{(3n^2-12n)/(2n^3-9n^2)}n2−n−122n2−15n+183n2−12n2n3−9n2?

1 Answer
Shwetank Mauria
Apr 3, 2016

(n(n+3)(2n-9))/(3(2n-3)(n-6))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))n2−n−122n2−15n+183n2−12n2n3−9n2 is equivalent to (n^2-n-12)/(2n^2-15n+18)xx(2n^3-9n^2)/(3n^2-12n)n2−n−122n2−15n+18×2n3−9n23n2−12n, 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)n2−n−12=(n2−4n+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)2n2−15n+18=2n2−12n−3n+18=2n(n−6)−3(n−6)=(2n−3)(n−6)

3n^2-12n=3n(n-4)3n2−12n=3n(n−4) and 2n^3-9n^2=n^2(2n-9)2n3−9n2=n2(2n−9)

Hence, (n^2-n-12)/(2n^2-15n+18)xx(2n^3-9n^2)/(3n^2-12n)n2−n−122n2−15n+18×2n3−9n23n2−12n 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))

Related questions
See all questions in Division of Rational Expressions
Impact of this question
1860 views around the world
You can reuse this answer
Creative Commons License