1.2.5: Dividing Polynomials
- Page ID
- 94948
By the end of this section, you will be able to:
- Reduce fractions
- Divide a monomial by a monomial
- Divide a polynomial by a monomial
- Divide polynomials using long division
Before you get started, take this readiness quiz.
1. Find the greatest common factor (GCF) of \(35\) and \(49\).
2. Reduce \(\dfrac{35}{49}\).
Reducing Fractions
Let's say we want to reduce the fraction \(\dfrac{60}{84}\). We need to find the largest integer that goes into \(60\) and \(84\) at the same time. Note that \(3\) goes into \(60\) and \(84\) since
\(60=3\cdot 20\qquad\text{and}\qquad 84 = 3\cdot 28.\)
So \(3\) is a common factor. However, \(4\) goes into \(20\) and \(28\) and so we can write
\(60=3\cdot (4\cdot 5)\qquad\text{and}\qquad 84 = 3\cdot (4\cdot 7)\)
or
\(60=(3\cdot 4)\cdot 5\qquad\text{and}\qquad 84 = (3\cdot 4)\cdot 7.\)
This means that \(3\cdot 4=12\) goes into \(60\) and \(84\) and we can write:
\( 60=12\cdot 5\qquad\text{and}\qquad 84 = 12\cdot 7.\)
So \(12\) is also a common factor of \(60\) and \(84\). Because \(5\) and \(12\) do not have any common factor other than \(1\), the greatest common factor of \(60\) and \(84\) is \(12\). Now we use the cancelation property:
\(\dfrac{a\cdot b}{a\cdot c}=\dfrac{b}{c}\)
to simplify \(\dfrac{60}{84}\):
\(\begin{align*}\dfrac{60}{84} &= \dfrac{12\cdot 5}{12\cdot 7}\\ &=\dfrac{5}{7}.\end{align*}\)
The reduction is \(\dfrac{5}{7}\). We want to keep the fraction format. The two fractions \(\dfrac{60}{84} \) and \(\dfrac{5}{7}\) are said to be equivalent as they represent the same number. If we had used the common factor \(3\) that we found in the beginning, we would have gotten
\(\begin{align*}\dfrac{60}{84} &= \dfrac{3\cdot 20}{3\cdot 28}\\ &=\dfrac{20}{28}\end{align*}\)
so that \(\dfrac{60}{84} \), \(\dfrac{5}{7}\) and \(\dfrac{20}{28}\) are all equivalent, but \(\dfrac{5}{7}\) is the one that cannot be simplified any further. This is the reduction we want.
Simplify:
a. \(\dfrac{36}{21}\)
b. \(\dfrac{72}{44}\)
c. \(\dfrac{8}{64}\)
Solution
a. The greatest common factor of \(36\) and \(21\) is \(3\).
\(\begin{align*} \dfrac{36}{21} & = \dfrac{3\cdot 12}{3\cdot 7}\\ & = \dfrac{12}{7}\end{align*} \)
b. The greatest common factor of \(72\) and \(44\) is \(4\).
\(\begin{align*} \dfrac{72}{44} & = \dfrac{4\cdot 18}{4\cdot 11}\\ & = \dfrac{18}{11}\end{align*} \)
c. The greatest common factor of \(8\) and \(64\) is \(8\).
\(\begin{align*} \dfrac{8}{64} & = \dfrac{8\cdot 1}{8\cdot 8}\\ & = \dfrac{1}{8}\end{align*} \)
Simplify:
a. \(\dfrac{108}{16}\)
b. \(\dfrac{25}{100}\)
c. \(\dfrac{81}{3}\)
- Answer
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a. \(\dfrac{27}{4}\)
b. \(\dfrac{1}{4}\)
c. \(27\)
Dividing Monomials
Now we will look at some examples where dividing two monomials results in a monomial (which is not always the case!!).
| Consider | \(\quad\dfrac{x^5}{x^2}\) |
|---|---|
| What do they mean? | \(=\dfrac{x\cdot x\cdot x\cdot x\cdot x}{x\cdot x}\) |
| Use the Equivalent Fractions Property. | \(=\dfrac{\cancel{x}\cdot\cancel{x}\cdot x\cdot x\cdot x}{\cancel{x}\cdot \cancel{x}}\) |
| Simplify. | \(=x^3\) |
| Note. |
\(\quad\dfrac{x^5}{x^2}=\dfrac{x^2x^3}{x^2}=x^3\) Reducing the fraction. |
Find the quotient \(54a^2b^3÷ (−6ab^2)\).
Solution
When we divide monomials with more than one variable, we write one fraction for each variable.
| \(\quad 54a^2b^3÷(−6ab^2)\) | |
|---|---|
| Rewrite as a fraction. | \(=\dfrac{54a^2b^3}{−6ab^2}\) |
| Use fraction multiplication. | \(=\dfrac{54}{−6}\cdot\dfrac{a^2}{a}\cdot\dfrac{b^3}{b^2}\) |
| Reduce the fraction. | \(=−9ab\) |
Find the quotient \(−72a^7b^3÷(8a^{5}b^2)\).
- Answer
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\(−{9}{a^2b}\)
Find the quotient \(−63c^8d^3÷(7c^{2}d)\).
- Answer
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\(−9c^6d^2\)
Once you become familiar with the process and have practiced it step by step several times, you may be able to simplify a fraction in one step.
Find the quotient \(\dfrac{14x^7y^{12}}{21x^{3}y^6}\).
Solution
Be very careful to simplify \(\dfrac{14}{21}\) by dividing out a common factor, and to simplify the variables by subtracting their exponents.
| \(\quad\dfrac{14x^7y^{12}}{21x^{3}y^6}\) | |
|---|---|
| Use fraction multiplication. | \(=\dfrac{14}{21}\cdot \dfrac{x^7}{x^3}\cdot\dfrac{y^{12}}{y^6}\) |
| Reduce. | \(=\dfrac{2}{3}x^4y^6\) |
Dividing a Polynomial by a Monomial
Do example with numbers here for distribution of division
Now that we know how to divide a monomial by a monomial, the next procedure is to divide a polynomial of two or more terms by a monomial. The method we’ll use to divide a polynomial by a monomial is based on the properties of fraction addition. So we’ll start with an example to review fraction addition.
The sum \(\dfrac{3}{7}+\dfrac{2}{7}=\dfrac{3+2}{7}\). So it is also true that \(\dfrac{3+2}{7}= \dfrac{3}{7}+\dfrac{2}{7}\), so you can distribute the divion by 7. You may also recall that division by 7 is the same as multiplication by \(\dfrac{1}{7}\) so, \(\dfrac{3+2}{7}=\dfrac{1}{7}(3+2)=\dfrac{1}{7}\cdot 3+\dfrac{1}{7}\cdot 2 =\dfrac{3}{7}+\dfrac{2}{7}\), or in words because you distribute multiplication over addition and subtraction, and division is multiplication is division by a reciprocal, you can also distribute division over addition and subtraction.
Here is another example with a variable.
The sum \(\dfrac{y}{5}+\dfrac{2}{5}\) simplifies to \(\dfrac{y+2}{5}\). Now we will do this in reverse to split a single fraction into separate fractions. For example, \(\dfrac{y+2}{5}\) can be written \(\dfrac{y}{5}+\dfrac{2}{5}\). Or, thinking of this as distibution: \(\dfrac{y+2}{5}=(y+2)\cdot\dfrac15=\dfrac{y}{5}+\dfrac25.\)
This is the “reverse” of fraction addition and it states that if a, b, and c are numbers where \(c\neq 0\), then \(\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}\). We will use this to divide polynomials by monomials.
To divide a polynomial by a monomial, divide each term of the polynomial by the monomial, or in other words, we distribute the division over addition and subtraction.
Find the quotient \((18x^3y−36xy^2)÷(−3xy)\).
Solution
| \(\quad (18x^3y−36xy^2)÷(−3xy)\) | |
| Rewrite as a fraction. | \(=\dfrac{18x^3y−36xy^2}{−3xy}\) |
| Divide each term by the divisor (distribute the division). Be careful with the signs! | \(=\dfrac{18x^3y}{−3xy}−\dfrac{36xy^2}{−3xy}\) |
| Simplify. | \(=−6x^2+12y\) |
Find the quotient \((32a^2b−16ab^2)÷(−8ab)\).
- Answer
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\(−4a+2b\)
Find the quotient \((−48a^8b^4−36a^6b^5)÷(−6a^3b^3)\).
- Answer
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\(8a^5b+6a^3b^2\)
We may, in certain situations, also divide a polynomial by a binomial as in the following example.
Find the quotient \((6(x-2)(3x-2))÷(3(x-2))\).
Solution
| \(\quad (6(x-2)(3x-2))÷(3(x-2))\) | |
| Rewrite as a fraction. | \(=\dfrac{6(x-2))(3x-2)}{3(x-2)}\) |
| Identify the common factors. | \(=\dfrac{3(x-2)2(3x-2)}{3(x-2)}\) |
| Simplify. | \(=2(3x-2)\) or \(6x-4\) |
Find the quotient \(((-4(2x-1))(2x+7))÷2(2x+7)\).
- Answer
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\(-2(2x-1)\)
Find the quotient \(((4(2x-1))(2x+7))÷-4(2x-1)\).
- Answer
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\(-(2x+7)\)
Dividing Polynomials Using Long Division
Divide a polynomial by a binomial, we follow a procedure very similar to long division of numbers. So let’s look carefully the steps we take when we divide a 3-digit number, 875, by a 2-digit number, 25.
\[\begin{array}{r} 35\phantom{5}\\ 25{\overline{\smash{\big)}\,875}}\\ \underline{-750}\\125\\\underline{-125}\\ 0\nonumber\end{array}\]
The quotient is \(35\) and the remainder is \(0\). We check division by multiplying the quotient by the divisor and then adding the remainder. If we did the division correctly, the result should equal the dividend.
\[35·25+0=875\checkmark\nonumber\]
Now we will divide a trinomial by a binomial. As you read through the example, notice how similar the steps are to the numerical example above.
Find the quotient and the remainder of \((x^2+9x+20)÷(x+5)\).
Solution
| \((x^2+9x+20)÷(x+5)\) | |
| Write it as a long division problem. Be sure the dividend is in standard form. |
\begin{array}{r} x+5{\overline{\smash{\big)}\,x^2+9x+20\phantom{)}}}\\ \nonumber\end{array} |
|
Divide \(x^2\) by \(x\). It may help to ask yourself, “What do I need Put the answer, \(x\), in the quotient over the \(x\) term. |
\begin{array}{r} x\phantom{x+20)}\\ x+5{\overline{\smash{\big)}\,x^2+9x+20\phantom{)}}}\\ \nonumber\end{array} |
Multiply \(x\) by \(x+5\). Change the sign of each term and put the answer under \(x^2+9x\). |
\begin{array}{r} x\phantom{(x+20)}\\ x+5{\overline{\smash{\big)}\,x^2+9x+20\phantom{)}}}\\ \underline{-x^2-5x\phantom{\;+20)}}\nonumber\end{array} |
| Add it to \(x^2+9x\). | \begin{array}{r} x\phantom{x+20)}\\ x+5{\overline{\smash{\big)}\,x^2+9x+20\phantom{)}}}\\ \underline{-x^2-5x\phantom{\;+20)}} \\ 4x+20\phantom{)}\\ \nonumber\end{array} |
Divide \(4x\) by \(x\). It may help to ask yourself, “What do I need to multiply \(x\) by to get \(4x\)?” Put the answer, \(4\), in the quotient over the constant term. |
\begin{array}{r} x+4\phantom{20)}\\ x+5{\overline{\smash{\big)}\,x^2+9x+20\phantom{)}}}\\ \underline{-x^2-5x\phantom{\;+20)}} \\4x+20\phantom{)}\\ \nonumber\end{array} |
| Multiply 4 by \(x+5\). Change the sign of each term and put the answer under \(4x+20\). | \begin{array}{r} x+4\phantom{20)}\\ x+5{\overline{\smash{\big)}\,x^2+9x+20\phantom{)}}}\\ \underline{-x^2-5x\phantom{\;+20)}} \\4x+20\phantom{)}\\ \underline{~\phantom{()}-4x-20} \nonumber\end{array} |
| Add it to \(4x+20\). |
\begin{array}{r} x+4\phantom{)}\\ x+5{\overline{\smash{\big)}\phantom{-} x^2+9x+20\phantom{)}}}\\ \underline{~\phantom{(}-x^2+(-5x)\phantom{\;+20)}}\\ 4x+20\phantom{)}\\ \underline{~\phantom{()}-4x-20}\\ 0 \phantom{)}\nonumber\end{array} |
|
Check.
|
\(\begin{array} {ll} {\text{Multiply the quotient by the divisor.}} &{(x+4)(x+5)} \\ {\text{You should get the dividend.}} &{x^2+9x+20\checkmark}\\ \end{array}\) |
| Conclude. | The quotient of \((x^2+9x+20)÷(x+5)\) is \(x+4\), and the remainder is \(0\). |
Find the quotient and the remainder of \((y^2+10y+21)÷(y+3)\).
- Answer
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The quotient is \(y+7\). The remainder is \(0\).
Find the quotient and the remainder of \((m^2+9m+20)÷(m+4)\).
- Answer
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The quotient is \(m+5\). The remainder is \(0\).
Look back at the dividends in previous examples. The terms were written in descending order of degrees, and there were no missing degrees. The dividend in this example will be \(x^4−x^2+5x−6\). It is missing an \(x^3\) term. We will add in \(0x^3\) as a placeholder.
Find the quotient and the remainder of \((x^4−x^2+5x−6)÷(x+2)\).
Solution
Notice that there is no \(x^3\) term in the dividend. We will add \(0x^3\) as a placeholder.
| \((x^4−x^2+5x−6)÷(x+2)\) | |
|---|---|
| Write it as a long division problem. Be sure the dividend is in standard form with placeholders for missing terms. | \begin{array}{r} x+2{\overline{\smash{\big)}\phantom{-} x^4+0x^3-x^2+5x-2\phantom{)}}}\nonumber\end{array} |
| Divide \(x^4\) by \(x\). Put the answer, \(x^3\), in the quotient over the \(x^3\) term. Multiply \(x^3\) by \(x+2\). Change the sign of each term and put the answer under \(x^4+0x^3\). Line up the like terms. Add them. |
\begin{array}{r} x^3\phantom{)-2x^2+3x-1)}\\ x+2{\overline{\smash{\big)}\phantom{-} x^4+0x^3-x^2+5x-2\phantom{)}}}\\ \underline{ -x^4-2x^3\phantom{\;-x^2+5x-2)}}\\ -2x^3-x^2+5x-2 \phantom{\;)}\nonumber\end{array} |
| Divide \(−2x^3\) by \(x\). Put the answer, \(−2x^2\), in the quotient over the \(x^2\) term. Multiply \(−2x^2\) times \(x+1\). Change the sign of each term and put the answer under Line up the like terms. Add them. |
\begin{array}{r} x^3-2x^2\phantom{))}\\ x+2{\overline{\smash{\big)}\phantom{-} x^4+0x^3-x^2+5x-2\phantom{)}}}\\ \underline{ -x^4-2x^3\phantom{\;-x^2+5x-2)}}\\ -2x^3-x^2+5x-2 \phantom{\;)}\\ \underline{2x^3+4x^2\phantom{\;+5x-2}}\\ 3x^2+5x-2 \phantom{\;)} \\ \nonumber\end{array}
|
| Divide \(3x^2\) by \(x\). Put the answer, \(3x\), in the quotient over the \(x\) term. Multiply \(3x\) times \(x+1\). Line up the like terms. Subtract and bring down the next term. |
\begin{array}{r} x^3-2x^2+3x\phantom{))}\\ x+2{\overline{\smash{\big)}\phantom{-} x^4+0x^3-x^2+5x-2\phantom{)}}}\\ \underline{ -x^4-2x^3\phantom{\;-x^2+5x-2)}}\\ -2x^3-x^2+5x-2 \phantom{\;)}\\ \underline{2x^3+4x^2\phantom{\;+5x-2}}\\ 3x^2+5x-2 \phantom{\;)} \\ \underline{~-3x^2-6x\phantom{)\;-2}}\\ -x-2\phantom{\;)} \\ \nonumber\end{array} |
| Divide \(−x\) by \(x\). Put the answer, \(−1\), in the quotient over the constant term. Multiply \(−1\) times \(x+1\). Line up the like terms. Change the signs, add. |
\begin{array}{r} x^3-2x^2+3x-1\phantom{))}\\ x+2{\overline{\smash{\big)}\phantom{-} x^4+0x^3-x^2+5x-2\phantom{)}}}\\ \underline{ -x^4-2x^3\phantom{\;-x^2+5x-2)}}\\ -2x^3-x^2+5x-2 \phantom{\;)}\\ \underline{2x^3+4x^2\phantom{\;+5x-2}}\\ 3x^2+5x-2 \phantom{\;)} \\ \underline{~-3x^2-6x\phantom{)\;-2}}\\ -x-2\phantom{\;)} \\ \underline{~x+2}\\ 0\phantom{\;)}\nonumber\end{array} |
| Check. | Multiply \((x+2)(x^3−2x^2+3x−1−4x+2)\). The result should be \(x^4−x^2+5x−6\). |
| Conclude. | The quotient of \((x^4−x^2+5x−6)÷(x+2)\) is \(x^3-2x^2+3x-1\), and the remainder is \(0\). |
Find the quotient and the remainder of \((x^4−7x^2+7x+6)÷(x+3)\).
- Answer
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The quotient is \(x^3−3x^2+2x+1+3x+3\). The remainder is \(0\).
Find the quotient and the remainder of \((x^4−11x^2−7x−6)÷(x+3)\).
- Answer
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The quotient is \(x^3−3x^2−2x−1−3x+3\). The remainder is \(0\).
In the next example, we will divide by \(2a−3\). As we divide, we will have to consider the constants as well as the variables.
Find the quotient and the remainder of \((8a^3+27)÷(2a+3)\).
Solution
This time we will show the division all in one step. We need to add two placeholders in order to divide.
\begin{array}{r} 4a^2-6a+9\phantom{))}\\ 2a+3{\overline{\smash{\big)}\phantom{-} 8a^3+0a^2+0a+27\phantom{)}}}\\ \underline{~ -8a^3-12a^2\phantom{\;+0a+27)}}\\ -12a^2+0a+27 \phantom{\;)}\\ \underline{12a^2+18a\phantom{\;+27}}\\ 18a+27 \phantom{\;)} \\ \underline{~-18a-27\phantom{)}}\\ 0\phantom{\;)} \nonumber\end{array}
To check, multiply \((2a+3)(4a^2−6a+9)\). The result should be \(8a^3+27\).
The quotient of \((8a^3+27)÷(2a+3)\) is \(4a^2-6a+9\) and the remainder is \(0\).
Find the quotient and the remainder of \((x^3−64)÷(x−4)\).
- Answer
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The quotient is \(x^2+4x+16\). The remainder is \(0\).
Find the quotient: \((125x^3−8)÷(5x−2)\).
- Answer
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The quotient is \(25x^2+10x+4\). The remainder is \(0\).
When we divided 875 by 25, we had remainder zero. But sometimes division of numbers does leave a remainder different from zero. The same is true when we divide polynomials. In the next example, we’ll have a division that leaves a remainder that is not zero. The degree of the remainder is always less than the degree of the divisor. To check, we need to verify that \((\text{quotient})(\text{divisor}) + \text{remainder} = \text{dividend}\).
Find the quotient and remainder when \(2x^3+3x^2+x+8\) is divided by \(x+2\).
Solution
| \((2x^3+3x^2+x+8)÷(x+2)\) | |
| Add it to \(4x+20\). |
\begin{array}{r} 2x^2-x+3\phantom{))}\\ x+2{\overline{\smash{\big)}\phantom{-} 2x^3+3x^2+x+8\phantom{)}}}\\ \underline{~\phantom{(}-2x^3-4x^2\phantom{\;+x+8)}}\\ -x^2+x+8 \phantom{\;)}\\ \underline{x^2+2x\phantom{+8}}\\ 3x+8 \phantom{\;)} \\ \underline{~\phantom{(}-3x-6\phantom{)}}\\ 2\phantom{\;)} \nonumber\end{array} |
|
Check.
|
\(\begin{align} (\text{quotient})(\text{divisor}) + \text{remainder} &= \text{dividend} \nonumber\\ (2x^2−1x+3)(x+2)+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3−x^2+3x+4x^2−2x+6+2 &\overset{?}{=} 2x^3+3x^2+x+8 \nonumber\\ 2x^3+3x^2+x+8 &= 2x^3+3x^2+x+8\checkmark \nonumber \end{align} \) |
| Conclude. | The quotient of \((2x^3+3x^2+x+8)÷(x+2)\) is \(2x^2-x+3\), and the remainder is \(2\). |
Find the quotient and remainder when \(3x^3+10x^2+6x−2\) is divided by \(x+2\).
- Answer
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The quotient is \(3x^2+4x−2\). The remainder is \(2\).
Find the quotient and remainder when \(4x^3+5x^2−5x+3\) is divided by \(x+2\).
- Answer
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The quotient is \(4x^2−3x+1\). The remainder is \(1\).
- Explain why you can distribute division over addition and subtraction.
- Can you divide a polynomial by a monomial and not get a polynomial? Give an example.
- Give another example of the type in the last exercise.
- What is the first step to reducing a fraction (numerator and denominator whole numbers)? Explain.
- Divide \((27x^3y^7-18x^7y^3+9x^2y^3)\div (9x^2y^3)\).
- Divide \(x^3-8x+3\) by \(x+5\) using long division.
Key Concepts
- Dividend, divisor, quotient, remainder
- Division of a polynomial by a monomial
- Long division of polynomials