SOLUTION: So I really have no clue what to do for this question. The textbook asks for the coefficient of x. My thinking was that the coefficient should be 16, however, the answer is 96. The

Algebra ->  Permutations -> SOLUTION: So I really have no clue what to do for this question. The textbook asks for the coefficient of x. My thinking was that the coefficient should be 16, however, the answer is 96. The      Log On


   

Question 1206241: So I really have no clue what to do for this question. The textbook asks for the coefficient of x. My thinking was that the coefficient should be 16, however, the answer is 96. The answer also calls for using Pascal's Triangle.
(2x+2)^4
Could someone please explain?

Found 2 solutions by Theo, greenestamps:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
your looking at (2x + 2) ^ 4

this is the same as (2x + 2) ^ 2 * (2x + 2) ^ 2

(2x + 2) ^ 2 equals:
(2x + 2) * (2x + 2) which equals:
2x * (2x + 2) + 2 * (2x + 2) which equals:
4x^2 + 4x + 4x + 4 which equals:
4x^2 + 8x + 4.

(4x^2 + 8x + 4) ^ 2 equals:
(4x^2 + 8x + 4) * (4x^2 + 8x + 4) which equals:
4x^2 * (4x^2 + 8x + 4) + 8x * (4x^2 + 8x + 4) + 4 * (4x^2 + 8x + 4).

4x^2 * (4x^2 + 8x + 4) = 16x^4 + 32x^3 + 16x^2.
8x * (4x^2 + 8x + 4) = 32x^3 + 64x^2 + 32x.
4 * (4x^2 + 8x + 4) = 16x^2 + 32x + 16.

total is 16x^4 + 32x^3 + 16x^2 + 32x^3 + 64x^2 + 32x + 16x^2 + 32x + 16.

combine like terms to get:

16x^4 + 64x^3 + 96x^2 + 64x + 16

i used an online binomial expansion calculator to confirm.
it gave me the same answer.
see below.



the correct answer is 96, but that's the coefficient of the x^2 term, not the x term.

here's my worksheet from the formula.



here's the calculator.

https://www.symbolab.com/solver/binomial-expansion-calculator/expand%20%5Cleft(2x%2B2%5Cright)%5E%7B4%7D?or=input


here's some references.

https://www.le.ac.uk/users/dsgp1/COURSES/MATHSTAT/2binome.pdf

https://www.math10.com/en/algebra/probabilities/binomial-theorem/binomial-theorem.html#:~:text=The%20Binomial%20Theorem%20Using%20Pascal's%20Triangle&text=(a%20%2B%20b)n%20%3D,st%20row%20of%20Pascal's%20triangle.



Answer by greenestamps(13367) About Me  (Show Source):
You can put this solution on YOUR website!


(NOTE: The answer is not 96....)

Here are the first several rows of Pascal's Triangle:

                   1  <------------- row 0
                 1   1  <----------- row 1
               1   2   1  <--------- row 2
             1   3   3   1  <------- row 3
           1   4   6   4   1  <----- row 4
         1   5  10  10   5   1  <--- row 5

The numbers in row n are the coefficients of the expansion of %28a%2Bb%29%5En.

For example, the "1 3 3 1" in row 3 are the coefficients in the expansion of

Your expression is 2x%2B2%29%5E4, so you are interested in the numbers in row 4 of Pascal's Triangle. Using your expression 2x%2B2%29 and the numbers in row 4....



You are asked for the coefficient of the x term. That term is %284%29%28%282x%29%5E1%29%282%29%5E3; the coefficient is %284%29%282%29%28%282%5E3%29%29=4%2A2%2A8=64.