The #x^3# term of the product will come from two places: multiplying the 1 from #1+bx# with the #x^3# term of #(x+2)^5# and multiplying the #bx# from #1+bx# by the #x^2# term of #(x+2)^5#.
From binomial expansions we can expand #(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#
So the squared term is #10(x)^2(2)^3=80x^2#.
The cubed term is #10(x)^3(2)^2=40x^3#.
For #(1+bx)(x+2)^5# the cubed term will be:
#(bx)(80x^2)+(1)(40x^3)=(80b+40)x^3#.
Since we want there to be no #x^3# term we need the coefficient, #80b+40# to equal 0.
#80b+40=0\rightarrowb=-1/2#.
