Binomial theorem

[pinklemonade]

So I have this question:
Find the coefficient of in the expansion of

The textbook shows me how to do it but its soo confusing and I really don't understand it at all! If anyone could explain it would be great!

[Tyleralp1]

I'm assuming there's a proper formula/methodology to do this, but just generically..

Using foil, just doing one bracket by one bracket...Expand the following:

(y+3)(y+3)(y+3)(2-y)(2-y)(2-y)(2-y)(2-y)

You'll end up with a pretty long equation to the 8th degree. But then you can simply find the co-efficient of y to the power of 4.

Yes, I know this is probably not the best way to do it, but just as a last resort haha

http://www.mathportal.org/calculators/polynomials-solvers/polynomials-expanding-calculator.php?val1=(yplus3)(yplus3)(yplus3)(2-y)(2-y)(2-y)(2-y)(2-y)&val2=%5Cleft(yplus3%5Cright)%7B%5Ccdot%7D%5Cleft(yplus3%5Cright)%7B%5Ccdot%7D%5Cleft(yplus3%5Cright)%7B%5Ccdot%7D%5Cleft(2-y%5Cright)%7B%5Ccdot%7D%5Cleft(2-y%5Cright)%7B%5Ccdot%7D%5Cleft(2-y%5Cright)%7B%5Ccdot%7D%5Cleft(2-y%5Cright)%7B%5Ccdot%7D%5Cleft(2-y%5Cright)

[brightsky]

This is a tricky question. You need to exercise a little logic. Expanding both binomials yields something of the form

(*y^3 + *y^2 + *y + *)(*y^5 + *y^4 + *y^3 + *y^2 + *y + *),

where * is some number. Now, ask yourself: if I expand the expression further, which terms, when multiplied together, will contribute to the coefficient in front of y^4 in the end result? Clearly, if we take the y^3 term from the expression enclosed in the first set of rounded brackets, and multiply it by the y term from the expression enclosed in the second set of rounded brackets, we will get a y^4 term. Other ways of obtaining a y^4 term are: multiplying the y^2 term from first expression by the y^2 term from the second expression, multiply the y term from the first expression by the y^3 term from the second expression, and multiplying the constant term from the first expression by the y^4 term from the second expression.

All that is left for us to do now is find the coefficients in front of the terms mentioned above. By applying the Binomial Theorem, whose statement is given on the following site: http://en.wikipedia.org/wiki/Binomial_theorem, you'll find that (y+3)^3 = y^3 + 9y^2 + 27 y + 27 and (2 - y)^5 = *y^5 + 10 y^4 - 40 y^3 + 80y^2 - 80y + *. I've blanked out the coefficients in front of the terms that were not mentioned above, since they are irrelevant for the purposes of this question. Now, recall that there are four different ways of forming the y^4 term by multiplying a term from the first expression by a term from the second expression. In order to find the coefficient in front of y^4 in the final expansion, all we need to do is multiply the coefficients in front of the terms that we are required to multiply together, and then add the four products. The working out is given below:

Coefficient in front of y^4 = 1*(-80) + 9*(80) + 27*(-40) + 27*(10) = -170.

Hope this makes sense!

[pinklemonade]

This is a tricky question. You need to exercise a little logic. Expanding both binomials yields something of the form

(*y^3 + *y^2 + *y + *)(*y^5 + *y^4 + *y^3 + *y^2 + *y + *),

where * is some number. Now, ask yourself: if I expand the expression further, which terms, when multiplied together, will contribute to the coefficient in front of y^4 in the end result? Clearly, if we take the y^3 term from the expression enclosed in the first set of rounded brackets, and multiply it by the y term from the expression enclosed in the second set of rounded brackets, we will get a y^4 term. Other ways of obtaining a y^4 term are: multiplying the y^2 term from first expression by the y^2 term from the second expression, multiply the y term from the first expression by the y^3 term from the second expression, and multiplying the constant term from the first expression by the y^4 term from the second expression.

All that is left for us to do now is find the coefficients in front of the terms mentioned above. By applying the Binomial Theorem, whose statement is given on the following site: http://en.wikipedia.org/wiki/Binomial_theorem, you'll find that (y+3)^3 = y^3 + 9y^2 + 27 y + 27 and (2 - y)^5 = *y^5 + 10 y^4 - 40 y^3 + 80y^2 - 80y + *. I've blanked out the coefficients in front of the terms that were not mentioned above, since they are irrelevant for the purposes of this question. Now, recall that there are four different ways of forming the y^4 term by multiplying a term from the first expression by a term from the second expression. In order to find the coefficient in front of y^4 in the final expansion, all we need to do is multiply the coefficients in front of the terms that we are required to multiply together, and then add the four products. The working out is given below:

Coefficient in front of y^4 = 1*(-80) + 9*(80) + 27*(-40) + 27*(10) = -170.

Hope this makes sense!

Thank you! Helped alot!

[chelonianlover]

Hey - I remember in yr 7 when I learned this theorem! (I know, I know... but I came across it on YouTube randomly, and got interested in it all. Don't judge LOL). Anyways, these videos alone helped me learn the theorem:

https://www.youtube.com/watch?v=1pSD8cYyqUo

https://www.youtube.com/watch?v=TeE-ypKj8ZI


I know you got the solution, but watching these may help your understanding in general.