+13 I think I need to use the Binomial Theorem here, but I don't know how.
Find the coefficient of y^4 in the expansion of (2y - 5 + y^2)^6.
expand (2y - 5 + y^2)^6=
y^12 + 12 y^11 + 30 y^10 - 140 y^9 - 585 y^8 + 792 y^7 + 4164 y^6 - 3960 y^5 - 14625 y^4 + 17500 y^3 + 18750 y^2 - 37500 y + 15625
y^4 = -14625
+118619 Thanks guest, I am just going to look at it a different way :)
Find the coefficient of y^4 in the expansion of (2y - 5 + y^2)^6.
\([(2y - 5) + y^2]^6\)
the general term will be \(\binom{6}{r}(y^2)^r(2y-5)^{6-r}\)
We are looking for the y^4 term so r can only be 0,1, or 2 (becasue after that I can see the y will have at least a power of 6
you need to do the three seperately and add the x^4 terms together to get the final answer.
I will do r =1, that is the hardest one anyway.
\(\binom{6}{r}(y^2)^r(2y-5)^{6-r}\)
\(\binom{6}{1}(y^2)(2y-5)^{5}\)
So now I need to find the coefficient of y^4 in \((2y-5)^{5}\)
That will be
\(\binom{5}{2}(2y)^2(-5)^3\\ =10*4*-125y^2\\ =-5000y^2\)
so the y^4 term will be
\(\binom{6}{1}(y^2)(-5000y^2)\\ =-30000y^4\)
Now you have to do y=0 and y=2 and add them all together.
