2.5: Dividing Polynomials
- Page ID
- 174511
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To succeed in this section, you'll need to use some skills from previous courses. While you should already know them, this is the first time they've been required. You can review these skills in CRC's Corequisite Codex. If you have a support class, it might cover some, but not all, of these topics.
The following is a list of learning objectives for this section.
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Lincoln Memorial, Washington, D.C. (credit: Ron Cogswell, Flickr)
The exterior of the Lincoln Memorial in Washington, D.C., is a large rectangular solid with length 61.5 meters (m), width 40 m, and height 30 m.1 We can easily find the volume using elementary Geometry.\[V=l \cdot w \cdot h =61.5 \cdot 40 \cdot 30 =73,800 \nonumber \]So the volume is 73,800 cubic meters (m3). Suppose we knew the volume, length, and width. We could divide to find the height.\[h= \dfrac{V}{l \cdot w} = \dfrac{73,800}{61.5 \cdot 40} =30. \nonumber \]As we can confirm from the dimensions above, the height is 30 m. We can use similar methods to find any of the missing dimensions. We can also use the same method if any or all of the measurements contain variable expressions. For example, suppose the volume of a rectangular solid is given by the polynomial \(3 x^4 - 3 x^3 -33 x^2 +54x\). The length of the solid is given by \(3x\) and the width is given by \(x-2\). To find the height of the solid, we can use polynomial division, which is the focus of this section.
Using Long Division to Divide Polynomials
Suppose we wish to find the zeros of \(f(x)=x^{3}+4 x^{2}-5 x-14 \). Setting \(f(x) = 0\) results in the polynomial equation \(x^{3}+4 x^{2}-5 x-14=0\). Despite all the factoring techniques we learned in Intermediate Algebra, this equation foils us at every turn. If we graph \(f\) using Desmos, we get the graph in Figure \( \PageIndex{2} \).
\( f(x) = x^3 + 4x^2 - 5x - 14 \)
The graph suggests that the function has three zeros, one of which is \(x = 2\). It's easy to show that \(f(2) = 0\), but the other two zeros seem less friendly. Even though we could click on the curve in Desmos to find decimal approximations for these, we seek an analytical method (a method by hand) to find the remaining zeros exactly.
Since \( f \) is a polynomial function and \( f(2) = 0 \), it should seem realistic that \( (x - 2) \) is a factor of \( f \). That is, if \(x = 2\) is a zero, it seems that there should be a factor of \((x − 2)\) lurking around in the factorization of \(f(x)\). Therefore, we should expect that \(x^{3}+4 x^{2}-5 x-14=(x-2) q(x)\), where \(q(x)\) is some other polynomial. If it even exists, how could we find such a \(q(x)\)? The answer comes from our old friend, polynomial division. Dividing \(x^{3}+4 x^{2}-5 x-14\) by \(x − 2\) gives\[ \require{enclose} \begin{array}{rl}
& \begin{array}{cccccccc} \quad & \quad & \quad & x^2 & + & 6 x & + & 7 \end{array} \\
x - 2 & \enclose{longdiv}{\begin{array}{cccccccc} \quad & x^3 & + & 4 x^2 & - & 5 x & - & 14 \end{array}} \\
& \underline{\begin{array}{cccccccc} - & (x^3 & - & 2x^2) & \quad & \quad & \quad & \quad \end{array} } \\
& \begin{array}{cccccccc} \quad & \quad & \quad & 6x^2 & - & 5x & \quad & \quad \end{array} \\
& \underline{\begin{array}{cccccccc} - & \quad & \quad & (6x^2 & - & 12x) & \quad & \quad \end{array}} \\
& \begin{array}{cccccccc} \quad & \quad & \quad & \quad & \quad & 7x & - & 14 \end{array} \\
& \underline{\begin{array}{cccccccc} - & \quad & \quad & \quad & \quad & (7x & - & 14) \end{array}}\\
& \begin{array}{cccccccc} \quad & \quad & \quad & \quad & \quad & \quad & \quad & 0 \end{array} \\
\end{array}\nonumber \]Thus, \( \frac{x^3 + 4x^2 - 5x - 14}{x - 2} = x^2 + 6x + 7 \). This means \(x^{3}+4 x^{2}-5 x-14=(x-2)\left(x^{2}+6 x+7\right)\), so to find the zeros of \(f\), we now solve \((x-2)\left(x^{2}+6 x+7\right)=0\). We get \(x − 2 = 0\) (which gives us our known zero, \(x = 2\)) as well as \(x^{2}+6 x+7=0\). The latter doesn't factor nicely, so we apply the Quadratic Formula to get \(x=-3 \pm \sqrt{2}\). The point of this section is to prepare us to be able to find as many zeros of a polynomial as possible by using division. We begin with a friendly reminder of what we can expect when we divide polynomials.
For example, we just performed long division and discovered that \( \frac{x^3 + 4x^2 - 5x - 14}{x - 2} = x^2 + 6x + 7 \), where the remainder is 0. In this equation, the dividend is \( x^3 + 4x^2 - 5x - 14 \), the divisor is \( x - 2 \), the quotient is \( x^2 + 6 x + 7 \), and (as was just mentioned) the remainder is 0.
Divide \(5 x^2 +3x-2\) by \(x+1\).
- Solution
-
\[ \require{enclose} \begin{array}{rl}
& \begin{array}{cccccc} \quad & \quad & \quad & 5 x & - & 2 \end{array} \\
x + 1 & \enclose{longdiv}{\begin{array}{cccccc} \quad & 5 x^2 & + & 3 x & - & 2 \end{array}} \\
& \underline{\begin{array}{cccccc} - & (5 x^2 & + & 5x) & \quad & \quad \end{array} } \\
& \begin{array}{cccccc} \quad & \quad & \quad & -2x & - & 2 \end{array} \\
& \underline{\begin{array}{cccccc} - & \quad & \quad & (-2x & - & 2) \end{array}} \\
& \begin{array}{cccccc} \quad & \quad & \quad & \quad & \quad & 0 \end{array} \\
\end{array}\nonumber \]The quotient is \(5x-2\). The remainder is 0. We write the result as\[\dfrac{5 x^2 +3x-2}{x+1} =5x-2 \nonumber \]or\[5 x^2 +3x-2=( x+1 )( 5x-2 ). \nonumber \]
The division problem in Example \( \PageIndex{ 1 } \) had a remainder of 0 and the result was that our original polynomial (the dividend) had a factor of \( (x + 1) \) (our divisor). This is a preview of an important theorem introduced in the next section.
Divide \(6 x^3 +11 x^2 -31x+15\) by \(3x-2\).
- Solution
-
\[ \require{enclose} \begin{array}{rl}
& \begin{array}{cccccccc} \quad & \quad & \quad & 2x^2 & + & 5 x & - & 7 \end{array} \\
3x - 2 & \enclose{longdiv}{\begin{array}{cccccccc} \quad & 6x^3 & + & 11 x^2 & - & 31 x & + & 15 \end{array}} \\
& \underline{\begin{array}{cccccccc} - & (6x^3 & - & 4x^2) & \quad & \quad & \quad & \quad \end{array} } \\
& \begin{array}{cccccccc} \quad & \quad & \quad & 15x^2 & - & 31x & \quad & \quad \end{array} \\
& \underline{\begin{array}{cccccccc} - & \quad & \quad & (15x^2 & - & 10x) & \quad & \quad \end{array}} \\
& \begin{array}{cccccccc} \quad & \quad & \quad & \quad & \quad & -21x & + & 15 \end{array} \\
& \underline{\begin{array}{cccccccc} - & \quad & \quad & \quad & \quad & (-21x & + & 14) \end{array}}\\
& \begin{array}{cccccccc} \quad & \quad & \quad & \quad & \quad & \quad & \quad & 1 \end{array} \\
\end{array}\nonumber \] There is a remainder of 1. We can express the result as:\[\dfrac{6 x^3 +11 x^2 -31x+15}{3x-2} =2 x^2 +5x-7+ \dfrac{1}{3x-2}. \nonumber \]We can check our work by using the Division Algorithm to rewrite the solution. Then multiply.\[(3x-2)(2 x^2 +5x-7)+1=6 x^3 +11 x^2 -31x+15. \nonumber \]Notice, as we write our result,- the dividend is \(6 x^3 +11 x^2 -31x+15\)
- the divisor is \(3x-2\)
- the quotient is \(2 x^2 +5x-7\)
- the remainder is \(1\)
Divide \(16 x^3 -12 x^2 +20x-3\) by \(4x+5\).
Focus on Calculus - Using Synthetic Division to Divide Polynomials
As we've seen, long division of polynomials can involve many steps and can be quite cumbersome. As such, we hunt for a better method. Fortunately, people like Ruffini and Horner have already blazed this trail. Synthetic division is a shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is 1 (although, it can be adjusted to divide by linear factors like \( 3x - 2) \)).
Let's look at the long division we performed at the beginning of the section and try to streamline it. First off, let's change all of the subtractions into additions by distributing through the negative signs.\[ \require{enclose} \begin{array}{rl}
& \begin{array}{ccccccc} \quad & \quad & x^2 & + & 6 x & + & 7 \end{array} \\
x - 2 & \enclose{longdiv}{\begin{array}{ccccccc} x^3 & + & 4 x^2 & - & 5 x & - & 14 \end{array}} \\
& \underline{\begin{array}{ccccccc} -x^3 & + & 2x^2 & \quad & \quad & \quad & \quad \end{array} } \\
& \begin{array}{ccccccc} \quad & \quad & 6x^2 & - & 5x & \quad & \quad \end{array} \\
& \underline{\begin{array}{ccccccc} \quad & \quad & -6x^2 & + & 12x & \quad & \quad \end{array}} \\
& \begin{array}{ccccccc} \quad & \quad & \quad & \quad & 7x & - & 14 \end{array} \\
& \underline{\begin{array}{ccccccc} \quad & \quad & \quad & \quad & -7x & + & 14 \end{array}}\\
& \begin{array}{ccccccc} \quad & \quad & \quad & \quad & \quad & \quad & 0 \end{array} \\
\end{array}\nonumber \]Next, observe that the terms \(-x^{3}\), \(-6 x^{2}\) and \(−7x\) are the exact opposite of the terms above them.\[ \require{enclose} \begin{array}{rl}
& \begin{array}{ccccccc} \quad & \quad & x^2 & + & 6 x & + & 7 \end{array} \\
x - 2 & \enclose{longdiv}{\begin{array}{ccccccc} x^3 & + & 4 x^2 & - & 5 x & - & 14 \end{array}} \\
& \underline{\begin{array}{ccccccc} \boxed{-x^3} & + & 2x^2 & \quad & \quad & \quad & \quad \end{array} } \\
& \begin{array}{ccccccc} \quad & \quad & 6x^2 & - & 5x & \quad & \quad \end{array} \\
& \underline{\begin{array}{ccccccc} \quad & \quad & \boxed{-6x^2} & + & 12x & \quad & \quad \end{array}} \\
& \begin{array}{ccccccc} \quad & \quad & \quad & \quad & 7x & - & 14 \end{array} \\
& \underline{\begin{array}{ccccccc} \quad & \quad & \quad & \quad & \boxed{-7x} & + & 14 \end{array}}\\
& \begin{array}{ccccccc} \quad & \quad & \quad & \quad & \quad & \quad & 0 \end{array} \\
\end{array}\nonumber \] Our algorithm ensures this is always the case, so we can omit them without losing any information. Also, note that the terms we "bring down" (namely the \(−5x\) and \(−14\)) aren't essential to recopy, so we omit them, too.\[ \require{enclose} \begin{array}{rl}
& \begin{array}{ccccccc} \quad & \quad & x^2 & + & 6 x & + & 7 \end{array} \\
x - 2 & \enclose{longdiv}{\begin{array}{ccccccc} x^3 & + & 4 x^2 & - & 5 x & - & 14 \end{array}} \\
& \underline{\begin{array}{ccccccc} \quad & \quad & 2x^2 & \quad & \quad & \quad & \quad \end{array} } \\
& \begin{array}{ccccccc} \quad & \quad & 6x^2 & \quad & \quad & \quad & \quad \end{array} \\
& \underline{\begin{array}{ccccccc} \quad & \quad & \quad & \quad & 12x & \quad & \quad \end{array}} \\
& \begin{array}{ccccccc} \quad & \quad & \quad & \quad & 7x & \quad & \quad \end{array} \\
& \underline{\begin{array}{ccccccc} \quad & \quad & \quad & \quad & \quad & \quad & 14 \end{array}}\\
& \begin{array}{ccccccc} \quad & \quad & \quad & \quad & \quad & \quad & 0 \end{array} \\
\end{array}\nonumber \]Now, let's move things up a bit and, for reasons that will become clear in a moment, copy the \(x^{3}\) into the last row.\[ \require{enclose} \begin{array}{rl}
& \begin{array}{ccccccc} \quad & \quad & x^2 & + & 6 x & + & 7 \end{array} \\
x - 2 & \enclose{longdiv}{\begin{array}{ccccccc} x^3 & + & 4 x^2 & - & 5 x & - & 14 \end{array}} \\
& \underline{\begin{array}{ccccccc} \quad & \quad & 2x^2 & \quad & 12x & \quad & 14 \end{array} } \\
& \begin{array}{ccccccc} x^3 & \quad & 6x^2 & \quad & 7x & \quad & 0 \end{array} \\
\end{array}\nonumber \]Note that by arranging things in this manner, each term in the last row is obtained by adding the two terms above it. Notice also that the quotient polynomial can be obtained by dividing each of the first three terms in the last row by \(x\) and adding the results. Take the time to work back through the original division problem. You will find that this is exactly the way we determined the quotient polynomial. This means we no longer need to write the quotient polynomial down, nor the \(x\) in the divisor, to determine our answer.\[\begin{array}{r}
\underline{-2} \mid & x^{3} & + & 4 x^{2} & - & 5 x & - & 14 \\
& & & 2 x^{2} & & 12 x & & 14 \\
\hline & x^{3} & & 6 x^{2} & & 7 x & & 0 \\
\end{array}\nonumber \]We've streamlined things quite a bit, but we can still do more. Let’s take a moment to remind ourselves where the \(2 x^{2}\), \(12x\) and \(14\) came from in the second row. Each of these terms was obtained by multiplying the terms in the quotient, \(x^{2}\), \(6x\) and \(7\), respectively, by the \(−2\) in \(x − 2\), then by \(−1\) when we changed the subtraction to addition. Multiplying by \(−2\) then by \(−1\) is the same as multiplying by \(2\), so we replace the \(−2\) in the divisor by \(2\). Furthermore, the coefficients of the quotient polynomial match the coefficients of the first three terms in the last row, so we now take the plunge and write only the coefficients of the terms to get\[\begin{array}{r}
\underline{2} \mid & 1 & 4 & -5 & -14 \\
& & 2 & 12 & 14 \\
\hline & 1 & 6 & 7 & 0 \\
\end{array}\nonumber \]We have constructed a synthetic division tableau for this polynomial division problem. Let's rework our division problem using this tableau to see how it greatly streamlines the division process. To divide \(x^{3}+4 x^{2}-5 x-14\) by \(x − 2\), we write \(2\) in the place of the divisor and the coefficients of \(x^{3}+4 x^{2}-5 x-14\) in for the dividend. Then, "bring down" the first coefficient of the dividend.\[\begin{array}{ccc}
\begin{array}{r}
\underline{2} \mid & 1 & 4 & -5 & -14 \\
\end{array} & \qquad \qquad \qquad & \begin{array}{r}
\underline{2} \mid & 1 & 4 & -5 & -14 \\
& \downarrow & & & \\
\hline & 1 & & & \\
\end{array} \end{array} \nonumber \]Next, take the \(2\) from the divisor and multiply by the \(1\) that was "brought down" to get \(2\). Write this underneath the \(4\), then add to get \(6\).\[\begin{array}{ccc}
\begin{array}{r}
\underline{2} \mid & 1 & 4 & -5 & -14 \\
& \downarrow & & & \\
\hline & 1 & & & \\
\end{array} & \qquad \qquad \qquad & \begin{array}{r}
\underline{2} \mid & 1 & 4 & -5 & -14 \\
& \downarrow & 2 & & \\
\hline & 1 & 6 & & \\
\end{array} \end{array} \nonumber \]Now take the \(2\) from the divisor times the \(6\) to get \(12\), and add it to the \(−5\) to get \(7\).\[\begin{array}{ccc}
\begin{array}{r}
\underline{2} \mid & 1 & 4 & -5 & -14 \\
& \downarrow & 2 & & \\
\hline & 1 & 6 & & \\
\end{array} & \qquad \qquad \qquad & \begin{array}{r}
\underline{2} \mid & 1 & 4 & -5 & -14 \\
& \downarrow & 2 & 12 & \\
\hline & 1 & 6 & 7 & \\
\end{array} \end{array} \nonumber \]Finally, take the \(2\) in the divisor times the \(7\) to get \(14\), and add it to the \(−14\) to get \(0\).\[\begin{array}{ccc}
\begin{array}{r}
\underline{2} \mid & 1 & 4 & -5 & -14 \\
& \downarrow & 2 & 12 & \\
\hline & 1 & 6 & 7 & \\
\end{array} & \qquad \qquad \qquad & \begin{array}{r}
\underline{2} \mid & 1 & 4 & -5 & -14 \\
& \downarrow & 2 & 12 & 14 \\
\hline & 1 & 6 & 7 & \boxed{0} \\
\end{array} \end{array} \nonumber \]The first three numbers in the last row of our tableau are the coefficients of the quotient polynomial. Remember, we started with a third-degree polynomial and divided by a first-degree polynomial, so the quotient is a second-degree polynomial. Hence the quotient is \(x^{2}+6 x+7\). The number in the box is the remainder.
Synthetic division is our tool of choice for dividing polynomials by divisors of the form \(x − c\). It is important to note that it works only for these kinds of divisors (you'll need to use good old-fashioned polynomial long division for divisors of degree larger than \(1\)). Also, take note that when a polynomial (of degree at least \(1\)) is divided by \(x − c\), the result will be a polynomial of exactly one less degree. Finally, it is worth the time to trace each step in synthetic division back to its corresponding step in long division. While the authors have done their best to indicate where the algorithm comes from, there is no substitute for working through it yourself.
- Write \(k\) for the divisor.
- Write the coefficients of the dividend.
- Bring the lead coefficient down.
- Multiply the lead coefficient by \(k\). Write the product in the next column.
- Add the terms of the second column.
- Multiply the result by \(k\). Write the product in the next column.
- Repeat steps 5 and 6 for the remaining columns.
- Use the bottom numbers to write the quotient. The number in the last column is the remainder and has degree 0, the next number from the right has degree 0, the next number from the right has degree 1, and so on.
Use synthetic division to divide \(5 x^2 -3x-36\) by \(x-3\).
- Solution
-
Begin by setting up the synthetic division. Write \(k\) and the coefficients.\[\begin{array}{r}
\underline{3} \mid & 5 & -3 & -36 \\
& & & \\
\hline & & & \\
\end{array}\nonumber \]Bring down the lead coefficient. Multiply the lead coefficient by \(k\).\[\begin{array}{r}
\underline{3} \mid & 5 & -3 & -36 \\
& & 15 & \\
\hline & 5 & & \\
\end{array}\nonumber \]Continue by adding the numbers in the second column. Multiply the resulting number by \(k\). Write the result in the next column. Then add the numbers in the third column.\[\begin{array}{r}
\underline{3} \mid & 5 & -3 & -36 \\
& & 15 & 36 \\
\hline & 5 & 12 & 0 \\
\end{array}\nonumber \]The result is \(5x+12\). The remainder is 0. So \(x-3\) is a factor of the original polynomial.Just as with long division, we can check our work by multiplying the quotient by the divisor and adding the remainder.\[(x-3)(5x+12)+0=5 x^2 -3x-36. \nonumber \]
Use synthetic division to divide \(4 x^3 +10 x^2 -6x-20\) by \(x+2\).
- Solution
-
The binomial divisor is \(x+2\) so \(k=-2\). Add each column, multiply the result by \( -2 \), and repeat until the last column is reached.\[\begin{array}{r}
\underline{-2} \mid & 4 & 10 & -6 & -20 \\
& & -8 & -4 & 20 \\
\hline & 4 & 2 & -10 & 0 \\
\end{array}\nonumber \]The result is \(4 x^2 +2x-10\). The remainder is 0. Thus, \(x+2\) is a factor of \(4 x^3 +10 x^2 -6x-20\).The graph of the polynomial function \(f(x)=4 x^3 +10 x^2 -6x-20\) in Figure \( \PageIndex{ 3 } \) shows a zero at \(x=k=-2\). This confirms that \(x+2\) is a factor of \(4 x^3 +10 x^2 -6x-20\).
Figure \( \PageIndex{ 3 } \)
Use synthetic division to divide \(-9 x^4 +10 x^3 +7 x^2 -6\) by \(x-1\).
- Solution
-
Notice there is no \( x \)-term. We will use a zero as the coefficient for that term.\[\begin{array}{r}
\underline{1} \mid & -9 & 10 & 7 & 0 & -6 \\
& & -9 & 1 & 8 & 8 \\
\hline & -9 & 1 & 8 & 8 & 2 \\
\end{array}\nonumber \]The result is \(-9 x^3 + x^2 +8x+8+ \frac{2}{x-1} \).
Use synthetic division to divide \(3 x^4 +18 x^3 -3x+40\) by \(x+7\).
Using Polynomial Division to Solve Application Problems
Polynomial division can be used to solve a variety of application problems involving expressions for area and volume. We looked at an application at the beginning of this section. Now we will solve that problem in the following example.
The volume of a rectangular solid is given by the polynomial \(3 x^4 -3 x^3 -33 x^2 +54x\). The length of the solid is given by \(3x\) and the width is given by \(x-2\). Find the height of the solid.
- Solution
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There are a few ways to approach this problem. We need to divide the expression for the volume of the solid by the expressions for the length and width. Let us create a sketch as in Figure \( \PageIndex{ 4 } \).
Figure \( \PageIndex{ 4 } \)
We can now write an equation by substituting the known values into the formula for the volume of a rectangular solid.\[ \begin{array}{rrcl}
& V & = & l \cdot w \cdot h \\[6pt]
\implies & 3 x^4 -3 x^3 -33 x^2 +54x & = & 3x \cdot (x-2) \cdot h \\[6pt]
\end{array} \nonumber \]To solve for \(h\), first divide both sides by \(3x\).\[ \begin{array}{rrcl}
& \dfrac{3x \cdot (x-2) \cdot h}{3x} & = & \dfrac{3 x^4 -3 x^3 -33 x^2 +54x}{3x} \\[6pt]
\implies & (x-2)h & = & x^3 - x^2 -11x+18 \\[6pt]
\end{array} \nonumber \]Now solve for \(h\) using synthetic division.\[h= \dfrac{x^3 - x^2 -11x+18}{x-2} \nonumber \]\[\begin{array}{r}
\underline{2} \mid & 1 & -1 & -11 & 18 \\
& & 2 & 2 & -18 \\
\hline & 1 & 1 & -9 & 0 \\
\end{array}\nonumber \]The quotient is \(x^2 +x-9\) and the remainder is 0. The height of the solid is \(x^2 +x-9\).
The area of a rectangle is given by \(3 x^3 +14 x^2 -23x+6\). The width of the rectangle is given by \(x+6\). Find an expression for the length of the rectangle.