Chapter 4 - Vector Spaces - 4.7 Change of Basis - Problems - Page 318: 8

$(-4,3,0)$

We know that $a(x^2+x)+b(2+2x)+c(1)=(-4x^2+2x+6)$. Thus $ax^2+ax+2b+2bx+c\\ =ax^2+x(a+2b)+(2b+c)\\ =-4x^2+2x+6$ Thus $a=-4 \\ a+2b=2 \\ 2b+c=6$ Thus $a=-4 \\ -4+2b=2 \\ 2b+c =6$ Thus $a=-4 \\ b= 3 \\ 2.3+c=6$ Thus $a=-4 \\ b=3 \\ c = 0$ Thus the vector is $(-4,3,0)$.