[tex] \frac{4v {}^{2} {y}^{5} }{3 {v}^{4} + 5v {}^{2} } [/tex]
your final answer
Answer:
4v^2y^5
-----------------------
(3u^4+5v^2)
Step-by-step explanation:
We can factor out a 2 v^2 out of the numerator and denominator
8v^4y^5 = (2v^2) *4v^2y^5
6u^4v^2 + 10 v^4 = 2v^2 (3u^4+5v^2)
So
8v^4y^5
-----------------
6u^4v^2 + 10 v^4
becomes
(2v^2) *4v^2y^5
2v^2 (3u^4+5v^2)
Canceling 2v^2
