Find Coefficient of X^5 in (3x^3 - 1/x^2)^10

[footprints]

Find the coefficient of the term X^5 of the expansion
[tex](3x^3 - \frac{1}{x^2})^{10}[/tex]

Another question off the topic.
Find the x-coordinate of the minimium point of [tex]y=2x^2-5x+3[/tex]
I know I have to complete the square but I'm not sure how its done.

 

[maverick280857]

Can you write the general term in the expansion of (x+y)^n? If you can, then you can replace x by 3x^3 and y by (-1/x^2).

For the second question, completing the square is a good idea if you do not know calculus (or are not supposed to use it).

Why not show your solution first?

 

[footprints]

I know calculus. So how do I do it using calculus?

 

[maverick280857]

Do you know the First and Second Derivative Tests?

What happens to a continuous function when its derivative switches sign? By a theorem called the Intermediate Value Theorem, every function which switches sign at least once over an interval must attain the value zero.

Try sketching a graph to convince yourself about the behavior of your quadratic polynomial.

At this point, you should consult your Calculus textbook for the First and Second Derivative Tests. If you have a problem, I'd be glad to help further.

Cheers
Vivek

 

[footprints]

Nope. Never heard of that.

 

[dextercioby]

Okay,forget about calculus.This is elementary.Take the previous advice to complete the square.

As for the first problem:The general term in the binomial expasion is
[tex] C_{n}^{k}a^{k}b^{n-k} [/tex]

Daniel.

 

[footprints]

But I forgot how to complete the square.

 

[dextercioby]

[tex] ax^{2}+bx+c=a(x^{2}+\frac{b}{a}x)+c=a[x^{2}+2\cdot (\frac{b}{2a})\cdot x+(\frac{b}{2a})^{2}]+c-a(\frac{b}{2a})^{2}=a(x+\frac{b}{2a})^{2}+c-\frac{b^{2}}{4a} [/tex]

Apply it.

Daniel.

 

[footprints]

Thanks for the help!

 

[dextercioby]

You're welcome.I hope you will master "completing the square",eventually...

Daniel.

P.S.It would be embarrasing to use calculus to find the maximum/minimum of a parabola...

 

[maverick280857]

P.S.It would be embarrasing to use calculus to find the maximum/minimum of a parabola...

I couldn't agree more...but you know its way faster than completing the square (you can write the answer by inspection and this is an asset if you're in a hurry). Nevertheless, its embarrasing .