Precalculus (6th Edition)

$\color{blue}{11x^2-x+2}$
Distribute $3x$ to obtain: $=x^2-3x+2-(x-4x^2)+3x(2x)+3x(1) \\=x^2-3x+2-(x-4x^2)+6x^2+3x$ Subtract each term of the binomial: $=x^3-3x+2-x-(-4x^2)+6x^2+3x \\=x^2-3x+2-x+4x^2+6x^2+3x$ Combine like terms to obtain: $=(x^2+4x^2+6x^2)+(-3x-x+3x)+2 \\=11x^2+(-x)+2 \\=\color{blue}{11x^2-x+2}$