Algebra

I'm sorry to say this but, think again guest answerer 

$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

(*ahhhh man not this again...)

Just know that:

$a=2y,\:\:b=-7$

when we are substituting values in the formula!!!

$=\sum _{i=0}^5\binom{5}{i}\left(2y\right)^{\left(5-i\right)}\left(-7\right)^i$

$=\frac{5!}{0!\left(5-0\right)!}\left(2y\right)^5\left(-7\right)^0+\frac{5!}{1!\left(5-1\right)!}\left(2y\right)^4\left(-7\right)^1+\frac{5!}{2!\left(5-2\right)!}\left(2y\right)^3\left(-7\right)^2+\frac{5!}{3!\left(5-3\right)!}\left(2y\right)^2\left(-7\right)^3+\frac{5!}{4!\left(5-4\right)!}\left(2y\right)^1\left(-7\right)^4+\frac{5!}{5!\left(5-5\right)!}\left(2y\right)^0\left(-7\right)^5$

$32y^5-560y^4+3920y^3-13720y^2+24010y-16807$

The coefficient of y^4 in the expansion of (2y-7)^5 is $-560$

-Vinculum

Edit: I double checked in Wolfram|Alpha.

It should be right.

https://www.wolframalpha.com/input?i2d=true&i=Power%5B%5Cleft%5C%2840%292y-7%5Cright%5C%2841%29%2C5%5D

Well, I did everything I can, I don't think I can solve this one then. Maybe ask someone or a mod !

-Vinculum