What is the coefficient of #x^7# in #(x^2+2x+3)^5#?
I understand how to find simple coefficients with #(x+y)^n# , but how do I approach this quadratic question?
I understand how to find simple coefficients with
1 Answer
Feb 19, 2018
Explanation:
It may be easiest to factor the quadratic and evaluate all of the terms of the powers of the binomials...
#(x^2+2x+3) = (x+1)(x+2)#
Then:
#(x+1)^5 = x^5+5x^4+10x^3+10x^2+5x+1#
Multiplying each successive term by successive powers of
#(x+2)^5 = x^5+10x^4+40x^3+80x^2+80x+32#
Then:
#(x^2+2x+3)^5 = (x+1)^5(x+2)^5#
which has term in
#x^5 * 80x^2 + 5x^4 * 40x^3 + 10x^3 * 10x^4 + 10x^2 * x^5#
#=(1*80+5*40+10*10+10*1) x^7#
#=(80+200+100+10) x^7#
#=390x^7#