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How do you find the coefficient of #x^2# in the expansion of #(x+2)^5#?

1 Answer
Oct 11, 2016

Answer:

The coefficient of the #x^2#-term is #80#.

Explanation:

First, we need to find the sixth row of Pascal's Triangle to determine the coefficient of the non-simplified terms of the expansion of the binomial expression. Remember that Pascal's Triangle begins with #1# as the first row and as the first and last entry of every other row. The middle terms of each row are obtained by adding the two terms from the row above.

                     1
                    / \
                  1    1
                 / \  / \
               1    2   1
              / \  / \  / \
             1   3   3   1
           /  \  / \  / \ / \
          1    4   6   4   1
        /  \  / \  / \  / \  / \
       1    5   10  10  5  1

Remember that in this case, #(x + 2)^5#, #a = x# and #b = 2#. The binomial theorem tells us that in the expansion, the terms will follow this pattern:

#a^5b^0 + a^4b^1 + a^3b^2 + a^2b^3 + a^1b^4 + a^0b^5#

Combining that with the coefficients from Pascal's Truangle gives us this expansion of #(x + 2)^5#:

#1(x^5)(2^0) + 5(x^4)(2^1) + 10(x^3)(2^2) + 10(x^2)(2^3) + 5(x^1)(2^4) + 1(x^0)(2^5)

The fourth term is the #x^2#-term, so we need to simplify it:
#10(x^2)(2^3) = 10(x^2)(8) = 80x^2

So, the coefficient of the #x^2#-term is #80#.