Give each child one cookie.
Now 10 children, 20 cookies.
So a+b+c+d+e+f+g+h+i+j=20
Result is (20+10-1) choose (10-1) = 29 C 9
= 10015005
Use the Binomial Theorem. What does this expansion look like in terms of $a$ and $n$? Can you set up some equations based on the given coefficients (-20 and 150 for x and x^2 respectively)?
Everyone, please do not answer this, it is an AoPS Algebra B writing problem. Ask on your message board. However if they are busy a helpful hint is that the sum of all coefficients is f(1) and the sum of the even coefficients minus the sum of odd coefficients is f(-1).
In both questions there are two angles congruent for the triangles that are in the question, so the desired results are each true by AA similarity.
Note that scaling the points up isn't going to change the area, so it's still 32.
I'm pretty sure there's a better way to do this. I've seen this question before. I have just forgotten how to approach it. Maybe let q(n)=p(2^n)-1/(2^n) for 0 <= n <=6 and then q(n)=n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)? Something like that? I could be wrong about the better way, though. It's just that 127/64 is a pretty nice number.
The roots can be (1,28) (2,14) (4,7) so that would mean 3.
