Question

I have little idea how to do part ii of this question, I've already done part 1 though
http://i.imgur.com/BgXvvmx.png
Thanks

Answer 1

after expanding the whole ordeal you will have a 7th order polynomial that will look like this

ax^7 + bx^6 + cx^5 + dx^4 + ex^3... etc

they are saying that the coefficient of x^3 [e in the general case above] = 0

you will have to find the value of k which will satisfy that.

let me know if this is unclear

Answer 2

I'm not exactly sure what you mean the answers say that k = 1
as \[\frac{ 270x^{3}}{ 4}-\frac{ 135kx^{3} }{ 2 }\]
I'm not that sure how they got to that term

Answer 3

they say that the coefficient of x^3 = 0. this means that.

\[\left( \frac{ 270 }{ 4 } - \frac{ 135k }{ 2 } \right) x^3 = 0\]or\[\frac{ 270 }{ 4 } = \frac{ 135k }{ 2 }\]
k = 1

Answer 4

Hmm ok I think I understand now thanks for your help!

Answer 5

glad i could help. if you have questions that could further your understanding lmk