SOLUTION: Find the coefficient of y^3 in the expansion of (y+3)^4 Thank you so much (^~^)

Question 1115996: Find the coefficient of y^3 in the expansion of (y+3)^4
Thank you so much (^~^)
Found 3 solutions by Boreal, josgarithmetic, greenestamps:
Answer by Boreal(15235)   (Show Source): You can put this solution on YOUR website!
The binomial expansion is 1 4 6 4 1
y^4+4y^3*3+6y^2*3^2+4y*3^3+3^4
The y^3 term has a coefficient of 12 ANSWER
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
Look at Pascal's Triangle, and the row which goes: 1, 4, 6, 4, 1. If you can determine where the y^3 term goes, then you know which value from the row to use, in order to find or determine what the coefficient is.
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

You have two responses from other tutors; both of them use the numbers 1 4 6 4 1 to find the answer. One tutor refers to the numbers as a row of Pascal's Triangle; the other refers to them as a binomial expansion.

And indeed the rows of Pascal's Triangle contain the coefficients of the binomial expansion of .

But let's take a look at WHY those are the numbers we want to use to solve this kind of problem. In other words, let's see WHY binomial expansion works the way it does.

If you understand why, the formula for binomial expansion will make sense; it will no longer seem like magic.

I will guess that you are familiar with the FOIL method for multiplying two binomials. The product of two binomials is the sum of four "partial products":
(1) F: the First terms in the two binomials,
(2) O: the Outer two terms,
(3) I: the Inner two terms, and
(4) L: the Last two terms.

That familiar method of multiplying two binomials is a simple case of a general principle which says that the product of any number of polynomials is the sum of all the "partial products" obtained by choosing one term from each polynomial.

In the expansion of , we are multiplying four identical factors of (y+3). The y^3 term in the expansion is the sum of all the partial products in which the "y" term is chosen in 3 of the 4 factors and the "3" is chosen in the other factor.

The number of ways of choosing 3 of the 4 factors from which to choose the "y" term is .

So the y^3 term is found by combining 4 partial products, in each of which the "y" term is used 3 times and the "3" is used once. That combining gives us the complete y^3 term in the expansion:

=

So the coefficient of the y^3 term in the expansion is 12.
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