What is the coefficient of #x^7# in #(x^2+2x+3)^5#?

Question

I understand how to find simple coefficients with #(x+y)^n#, but how do I approach this quadratic question?

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Answer 1 · 2018-02-19T21:42:01

#390#

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 #2#, we find:

#(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^7# formed by multiplying combinations of terms:

#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#