Thực hiện phép chia ta thấy :
\(\left(7x^5-3x^3+x\right):5x=\frac{7}{5}x^4-\frac{3}{5}x^2+\frac{1}{5}\)
\(\left(4x^2+4x+1\right):\left(2x+1\right)=\left(2x+1\right)^2:\left(2x+1\right)=2x+1\)
\(\left(x^4+x^3-x^2+2x\right):\left(x^2-x+1\right)=x^2+2x\)
\(\left(4x^2-4x+3\right):\left(2x-1\right)=\left[\left(2x-1\right)^2+2\right]:\left(2x-1\right)\) $=2x - 1$ và dư $2$.