How do you find the coefficient of x^5 in the expansion of (2x+3)(x+1)^8?

1 Answer
George C.
Jul 12, 2015

Calculate the coefficients of #x^5# and #x^4# in #(x+1)^8#, then multiply by #3# and #2# then add to get the answer #308#

Explanation:

#(x+1)^8 = sum_(n=0)^(n=8) ((8),(n))x^n#

The coefficient of #x^4# in #(x+1)^8# is #((8),(4)) = (8!)/(4!4!) = 70#

The coefficient of #x^5# in #(x+1)^8# is #((8),(5)) = (8!)/(5!3!) = 56#

So the coefficient of #x^5# in #(2x+3)(x+1)^8# is

#2*70 + 3*56 = 140 + 168 = 308#

Related questions
See all questions in The Binomial Theorem
Impact of this question
28097 views around the world
You can reuse this answer
Creative Commons License