Cho ba đa thức A = 4x⁴ – 2 + 5x² – x; B = 5x + 3 – 4x² – 3x³ và C = 4x⁴ + 4x – 4x³ + x².

Sắp xếp các đa thức trên theo lũy thừa giảm của biến.

(x² - 3x + 2) + (4x³ - x² + x - 1) = x² - 3x + 2 + 4x³ - x² + x - 1

= 4x³ + (x² - x²) + (-3x + x) + (2 - 1)

= 4x³ - 2x + 1.

Cách thứ nhất:

\[\begin{array}{l} + \underline {\begin{array}{*{20}{c}}\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6{x^4} - 4{x^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + x - \frac{1}{3}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 3{x^4} - 2{x^3} - 5{x^2} + x + \frac{2}{3}\end{array}\end{array}} \\{\rm{A}}\,\,{\rm{ + }}\,\,{\rm{B = }}\,\,{\rm{3}}{x^4} - {\rm{6}}{{\rm{x}}^3} - 5{x^2} + 2x + \frac{1}{3}\end{array}\]

\[\begin{array}{l} - \underline {\begin{array}{*{20}{c}}\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6{x^4} - 4{x^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + x - \frac{1}{3}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 3{x^4} - 2{x^3} - 5{x^2} + x + \frac{2}{3}\end{array}\end{array}} \\{\rm{A }} - {\rm{ B = 9}}{x^4} - 2{{\rm{x}}^3} + 5{x^2}\,\,\,\,\,\,\,\,\,\, - 1\end{array}\]

Cách thứ hai:

A + B = (6x⁴ - 4x³ + x - \[\frac{1}{3}\]) + (-3x⁴ - 2x³ - 5x² + x + \[\frac{2}{3}\])

= (6x⁴ - 3x⁴) + (-4x³ - 2x³) - 5x² + (x + x) + \[\left( { - \frac{1}{3} + \frac{2}{3}} \right)\]

= 3x⁴ - 6x³ - 5x² + 2x + \[\frac{1}{3}\]

A - B = (6x⁴ - 4x³ + x - \[\frac{1}{3}\]) - (-3x⁴ - 2x³ - 5x² + x + \[\frac{2}{3}\])

= 6x⁴ - 4x³ + x - \[\frac{1}{3}\] + 3x⁴ + 2x³ + 5x² - x - \[\frac{2}{3}\]

= (6x⁴ + 3x⁴) + (-4x³ + 2x³) + 5x² + (x - x) + \[\left( { - \frac{1}{3} - \frac{2}{3}} \right)\]

= 9x⁴ - 2x³ + 5x² - 1