Question

divide assume the denominator does not equal zero
(18x^5y^3 + 6x^2y^4 + 24x^6y) ÷ 3xy

Answer 1

my advise it to split it into smaller problems

\[\frac{18x^5y^3}{3xy} + \frac{6x^2y^4}{3xy} + \frac{24x^6y}{3xy}\]

then approach it term by term

the 1st fraction can be simplified to

\[6x^4y^2 + \frac{6x^2y^4}{3xy} + \frac{24x^6y}{3xy}\]

you'll find 3xy divides evenly into each