How do you divide #\frac { 5n ^ { 4} + 40n ^ { 3} } { n ^ { 2} - 19n + 48} \div \frac { 3n ^ { 2} + 15n - 72} { n ^ { 2} - 16n }#?

Question

599 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-21T01:01:47

Convert to a multiplication.

#=(5n^4 + 40n^3)/(n^2 - 19n + 48) * (n^2 - 16n)/(3n^2 + 15n - 72)#

Factor everything.

#=(5n^3(n + 8))/((n - 16)(n - 3)) * (n(n - 16))/(3(n^2 + 5n - 24))#

#=(5n^3(n + 8))/((n - 16)(n - 3)) * (n(n - 16))/(3(n + 8)(n - 3))#

After eliminating, we are left with:

#=(5n^4)/(3(n + 3)^2)#

With restrictions of:

#n != 16, 3, -8, 3 and 0#

Hopefully this helps!