Chapter 4 - Vector Spaces - 4.7 Change of Basis - Problems - Page 318: 10

$(5,-2,3,-2)$

We know that $a(1)+b(1+x)+c(1+x+x^2)+d(1+x+x^2+x^3)=4-x+x^2-2x^3$. Thus $a+b+bx+c+cx+cx^2+d+dx+dx^2+dx^3\\ =(a+b+c+d)+(b+c+d)x+(c+d)x^2+dx^3\\ =4-x+x^2-2x^3$ Thus $a+b+c+d=4\\ b+c+d=-1 \\ c+d=1 \\ d=-2$ Thus $a+b+c+d=4\\ b+c+d=-1 \\ c=3 \\ d=-2$ Thus $a+b+c+d=4\\ b= -2 \\ c=3 \\ d=-2$ Thus $a=5 \\ b=-2 \\ c=3 \\ d=-2$ Thus the vector is $(5,-2,3,-2)$.