Algebra II expand correct....?

Question

Thanx to someone answering my earlier question i know how to do expand. At least i think. Is this long assed answer correct? (a+b)^14=a^14+14a^13b+101a^12b^2+414a^11b^3+1101a^10b^4+2103a^9b^5+3053a^8b^6+3452a^7b^7+3053a^6b^8+2103a^5b^9+1101a^4b^10
+414a^3b^11+101a^2b^12+14ab^13+b^14
Next question.... I think i understand this correct me if wrong: (2x+y)^4=2x^4+8x^3y+12x^2y^2+8xy^3+y^4 I don't think it is 2x^4+8x^3y+12x^2y^2+8xy^3+2y^4 but if it is tell me. Please grade me -0=2:2, -1=1:2,
-2=0:2. If i got any wrong please tell me what i did?

with (2x+y)^4 I used 1 4 6 4 1 but then i thought since it is 2x i should use 2 8 12 8 2. But you said the whole (2x+y)^4= 16x+y^4? Do I use 16 64 96 64 16? Answer: 16x^4 + 64x^3y + 96x^2y^2 +64xy^3 + 16y^4? I think I'm doomed ;( and Doug i don't understand was the first answer wrong?

Ahhh My pascal's triangle got messed up no! But i fixed it and created another so i should be fine. 15 layers is alot to a triangle though all hand written ;(

Answer 1

You're on the path to righteous, Buddy, but you still gotta keep better track of those terms. Let's see if I can help:

(a+b)^n = sum of (nCr)*a^(n-r)*b^r where r goes from 0 to n and (nCr) is the 'binomial function' n!/(r!*(n-r)!) with the ! meaning factorial (n! means n*(n-1)*(n-2)*...2*1 so 4! = 4*3*2*1 = 24 and 0! = 1 by definition)

That looks scary, but it's actually one *helluva* lot easier than doing 14 polynomial multiplications.

First term (r=0) (nCr) = 14!/(0!*(14!)) = 1
a^(14-0) = a^14
b^0 = 1
so 1'st term is a^14

2'nd term (r=1) (nCr) = 14!/(1!*(13!)) = 14
a*(14-1) = a^13
b^1
so 2,nd term is 14*(a^13)*b

3'rd term (r=2) (nCr) = 14!/(2!*(12!)) = 91 (not 101)
a^(14-2) = a^12
b^2
so 3'rd term is 91*(a^12)*b^2

4'th term (r=3) (nCr) = 14!/(3!*(11!)) = 364
a^(14-3) = a^11
b^3
so 4'th term is 364*(a^11)*b^3
and so on.

The (nCr) function is *so* important in math that, especially on scientific calculators, there's frequently a special button to calculate it. If yours has one, you have it made. If not, at least see if it has a factorial (n!) key. If not, spend 10 bucks at Walgreens or K-Mart and get one that does

Also, anonymous was totally correct about (2x+y)^n You have to use the entire 2x as the a in (a+b)^n.

Hang in there, Dewd. You're studying and thinking and that's *always* the right thing.


Doug

Answer 2

(a + b)^n = ∑ nCk · a^(n - k) · b^(k) , for k going from 0 to n.

Problem 1:
(a + b)^14 = a^14 + 14a^13b + 91a^12b^2 + 364a^11b^3 + 1001a^10b^4 + 2002a^9b^5 + 3003a^8b^6 + 3432a^7b^7 + 3003a^6b^8 + 2002a^5b^9 + 1001a^4b^10 + 364a^3b^11 + 91a^2b^12 + 14ab^13 + b^14
Your variables are right on, but your coefficients are off... you're messing up somewhere in your combinations.
nCk = n! / [k! (n - k)!]

Problem 2:
(2x + y)^4
= 1(2x)^4y^0 + 4(2x)^3y + 6(2x)^2y^2 + 4(2x)^1y^3 + 1(2x)^0y^4
= 1 · 16x^4 + 4 · 8x^3y + 6 · 4x^2y^2 + 4 · 2xy^3 + 1y^4
= 16x^4 + 32x^3 · y + 24x^2 · y^2 + 8x · y^3 + y^4
Your coefficients are again off, this times because you weren't taking the (2) in the (2x) into account when raising to the 4th, 3rd, and 2nd powers.

Answer 3

I think the first one is right, but the second one is definitely not. You have to raise the whole term (including the coefficient) to the right power in each term. So the first term is (2x)^4, or 16x^4, and similarly for the rest of the terms. It's been a while since I did this, but I can tell that the second one is wrong from looking at it. Also, next time add some spaces between the terms because the information gets cut off if you don't (we can't see the majority of the stuff you typed the way it is). Good luck!

Answer 4

(2x + y)(2x + y)(2x + y)(2x + y)

(4x^2 + 2xy + 2xy + y^2)(4x^2 + 2xy + 2xy + y^2)

(4x^2 + 4xy + y^2)(4x^2 + 4xy + y^2)

16x^2 + 16x^3y + 4x^2y^2 + 16x^3y + 16x^2y^2 + 4xy^3 + 4x^2y^2 + 4xy^3 + y^4

16x^2 + (16 + 16)x^3y + (16 + 4 + 4)x^2y^2 + (4 + 4)xy^3 + y^4

16x^2 + 32x^3y + 24x^2y^2 + 8xy^3 + y^4

so

(2x + y)^4 = 16x^4 + 32x^3y + 24x^2y^2 + 8xy^3 + y^4

www.quickmath.com will also expand your problem for you

just click on Expand under Algebra, then type in

(a + b)^14, click expand
(2x + y)^4, click expand

just make sure to write the variables in the right order.