By now we have seen how to add and subtract polynomials by collecting like terms and multiplying polynomials using distributive property . Now, we will learn how to divde a polynomial by a monomial. Remember, a monomial is a polynomial with only one term. That means no addition or subtraction. Dividing by polynomials is a little more complicated so we will save that for later.
Simplify the expression \( \cfrac{15x^5 - 10x^4 + 5x^3}{5x^2}\).
To divide by a monomial, we will divide each term in the numerator by the monomial.
\( \cfrac{15x^5 - 10x^4 + 5x^3}{\textcolor{red}{5x^2}} \)
\( =\cfrac{15x^5}{\textcolor{red}{5x^2}} + \cfrac{- 10x^4}{\textcolor{red}{5x^2}} + \cfrac{5x^3}{\textcolor{red}{5x^2}}\)
You can think of this like the distributive property. We are distributing \(\cfrac{1}{5x^2}\) into the numerator.
Next, we can simplify each term. First, we can simplify the coefficients:
\( =\cfrac{\textcolor{red}{15}x^5}{\textcolor{red}{5}x^2} + \cfrac{\textcolor{red}{-10}x^4}{\textcolor{red}{5}x^2} + \cfrac{\textcolor{red}{5}x^3}{\textcolor{red}{5}x^2}\)
\( =\cfrac{\cancel{\textcolor{red}{15}}3x^5}{\cancel{\textcolor{red}{5}}x^2} + \cfrac{-\cancel{\textcolor{red}{10}}2x^4}{{\cancel{\textcolor{red}{5}}x^2}} + \cfrac{\cancel{\textcolor{red}{5}}x^3}{{\cancel{\textcolor{red}{5}}x^2}}\)
\( =\cfrac{3x^5}{x^2} + \cfrac{-2x^4}{x^2} + \cfrac{x^3}{x^2}\)
Then, we can simplify the exponents! Remember the exponent rules for dividing exponents with the same base:
\( =\cfrac{3\textcolor{red}{x}^\textcolor{green}{5}}{\textcolor{red}{x}^\textcolor{green}{2}} + \cfrac{-2\textcolor{red}{x}^\textcolor{green}{4}}{\textcolor{red}{x}^\textcolor{green}{2}} + \cfrac{\textcolor{red}{x}^\textcolor{green}{3}}{\textcolor{red}{x}^\textcolor{green}{2}}\)
\(= 3\textcolor{red}{x}^{\textcolor{green}{5-2}} + -2\textcolor{red}{x}^{\textcolor{green}{4-2}} + \textcolor{red}{x}^{\textcolor{green}{3-2}}\)
\( =3x^3 + -2x^2 + x^1\)
\( =3x^3 + -2x^2 + x\)
Therefore, we can determine that the expression simplified is \(\boldsymbol{3x^3 + -2x^2 + x}\).
\(\cfrac{4x^2y + 12xy -8xy^2}{2xy}\)
First, divide each term in the numerator by the denominator:
\(\cfrac{4x^2y + 12xy -8xy^2}{\textcolor{red}{2xy}}\)
\( =\cfrac{4x^2y}{\textcolor{red}{2xy}} + \cfrac{12xy}{\textcolor{red}{2xy}} + \cfrac{-8xy^2}{\textcolor{red}{2xy}}\)
Next, simplify the coefficients:
\( =\cfrac{\textcolor{red}{4}x^2y}{\textcolor{red}{2}xy} + \cfrac{\textcolor{red}{12}xy}{\textcolor{red}{2}xy} + \cfrac{\textcolor{red}{-8}xy^2}{\textcolor{red}{2}xy}\)
\( =\cfrac{\cancel{\textcolor{red}{4}}2x^2y}{\cancel{\textcolor{red}{2}}xy} + \cfrac{\cancel{\textcolor{red}{12}}6xy}{\cancel{\textcolor{red}{2}}xy} + \cfrac{\cancel{\textcolor{red}{-84}}xy^2}{\cancel{\textcolor{red}{2}}xy}\)
\( =\cfrac{2x^2y}{xy} + \cfrac{6xy}{xy} + \cfrac{-4xy^2}{xy}\)
Next, we can simplify the exponents for \(x\):
\( =\cfrac{2\textcolor{red}{x}^\textcolor{green}{2}y}{\textcolor{red}{x}^\textcolor{green}{1}y} + \cfrac{6\textcolor{red}{x}^\textcolor{green}{1}y}{\textcolor{red}{x}^\textcolor{green}{1}y} + \cfrac{-4\textcolor{red}{x}^\textcolor{green}{1}y^2}{\textcolor{red}{x}^\textcolor{green}{1}y}\)
\( =\cfrac{2\textcolor{red}{x}^{\textcolor{green}{2 - 1}}y}{y} + \cfrac{6\textcolor{red}{x}^{\textcolor{green}{1-1}}y}{y} + \cfrac{-4\textcolor{red}{x}^{\textcolor{green}{1-1}}y^2}{y}\)
\( =\cfrac{2xy}{y} + \cfrac{6y}{y} + \cfrac{-4y^2}{y}\)
After, we can simplify the exponents for \(y\):
\( =\cfrac{2x\textcolor{red}{y}^\textcolor{green}{1}}{\textcolor{red}{y}^\textcolor{green}{1}} + \cfrac{6\textcolor{red}{y}^\textcolor{green}{1}}{\textcolor{red}{y}^\textcolor{green}{1}} + \cfrac{-4\textcolor{red}{y}^\textcolor{green}{2}}{\textcolor{red}{y}^\textcolor{green}{1}}\)
\(= 2x\textcolor{red}{y}^{\textcolor{green}{1-1}} + 6\textcolor{red}{y}^{\textcolor{green}{1-1}}-4\textcolor{red}{y}^{\textcolor{green}{2-1}}\)
\(= 2x + 6-4y\)
Finally, we can rearrange the terms:
\(= 2x - 4y + 6\)
Therefore, we can determine that the expression simplified is \(\boldsymbol{2x - 4y + 6}\).
Sometimes, the exponents on the denominator will be larger than the exponent on the numberator. For these scenarios, leave the term in the denominator with a positive exponent:
\( \cfrac{12xy^3}{-2x^2y^2} \)
\(= \cfrac{\textcolor{red}{12}\textcolor{green}{x}\textcolor{brown}{y^3}}{\textcolor{red}{-2}\textcolor{green}{x^2}\textcolor{brown}{y^2}} \)
\(= \cfrac{\textcolor{red}{-6}\textcolor{brown}{y}}{\textcolor{green}{x}} \)