5.4: Dividing Polynomials (2025)

\[\require{enclose}\begin{array} {rll} \largex+2\enclose{longdiv}{ 2x^3-3x^2+4x+5\phantom{0}} &\qquad &\large\text{Set up the division problem.}\\[8pt]
\large2x^2 \hspace{5.45em}&\qquad & \large2x^3\text{ divided by }x\text{ is }2x^2.\\[-3pt]
\largex+2\enclose{longdiv}{ 2x^3-3x^2+4x+5\phantom{0}} &\qquad &\\[8pt]
\large2x^2 \hspace{5.45em}&\qquad & \\[-3pt]
\largex+2\enclose{longdiv}{ 2x^3-3x^2+4x+5\phantom{0}} &\qquad &\\[-3pt]
\large \underline{-\left( 2x^3 + 4x^2 \right)}\hspace{4.9em}&\qquad & \large\text{Multiply }x+2\text{by}2x^2\text{ and subtract.}\\[-3pt]
\large -7x^2+4x \hspace{2.8em}&\qquad & \large\text{Bring down the next term.}\\[8pt]

\large2x^2 - 7x\hspace{2.85em}&\qquad & \large -7x^2\text{ divided by }x\text{ is }-7x.\\[-3pt]
\largex+2\enclose{longdiv}{ 2x^3-3x^2+4x+5\phantom{0}} &\qquad &\\[-3pt]
\large \underline{-\left( 2x^3 + 4x^2 \right)}\hspace{4.9em}&\qquad & \\[-3pt]
\large -7x^2+4x \hspace{2.8em}&\qquad & \\[-3pt]
\large \underline{-\left( -7x^2-14x\right)}\hspace{2.0em}&\qquad & \large\text{Multiply }x+2\text{by}-7x.\\[-3pt]
\large 18x+5\phantom{0}&\qquad & \large\text{Subtract and bring down the next term.}\\[8pt]

\large2x^2 - 7x+18&\qquad & \large 18x\text{ divided by }x\text{ is }18.\\[-3pt]
\largex+2\enclose{longdiv}{ 2x^3-3x^2+4x+5\phantom{0}} &\qquad &\\[-3pt]
\large \underline{-\left( 2x^3 + 4x^2 \right)}\hspace{4.9em}&\qquad & \\[-3pt]
\large -7x^2+4x \hspace{2.8em}&\qquad & \\[-3pt]
\large \underline{-\left( -7x^2-14x\right)}\hspace{2.0em}&\qquad & \\[-3pt]
\large 18x+\phantom{0}5&\qquad & \\[-3pt]
\large \underline{-\left( 18x+ 36\right)} \hspace{-0.45em}&\qquad & \large\text{Multiply }x+2\text{by}18.\\[-3pt]
\large -31&\qquad & \large\text{Subtract.}\\[8pt]
\end{array} \nonumber\]

The Division Algorithm

The Division Algorithm states that, given a polynomial dividend \(f(x)\) and a non-zero polynomial divisor \(d(x)\) where the degree of \(d(x)\) is less than or equal to the degree of \(f(x)\), there exist unique polynomials \(q(x)\) and \(r(x)\) such that

\[f(x)=d(x)q(x)+r(x)\]

\(q(x)\) is the quotient and \(r(x)\) is the remainder. The remainder is either equal to zero or has degree strictly less than \(d(x)\).

If \(r(x)=0\), then \(d(x)\) divides evenly into \(f(x)\). This means that, in this case, both \(d(x)\) and \(q(x)\) are factors of \(f(x)\).

Given a polynomial and a binomial, use long division to divide the polynomial by the binomial

1. Set up the division problem.
2. Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor.
3. Multiply the answer by the divisor and write it below the like terms of the dividend.
4. Subtract the bottom binomial from the top binomial.
5. Bring down the next term of the dividend.
6. Repeat steps 2–5 until reaching the last term of the dividend.
7. If the remainder is non-zero, express as a fraction using the divisor as the denominator.

Example \(\PageIndex{1}\): Using Long Division to Divide a Second-Degree Polynomial

Divide \(5x^2+3x−2\) by \(x+1\).

Solution

The quotient is \(5x−2\). The remainder is 0. We write the result as

\[\dfrac{5x^2+3x−2}{x+1}=5x−2 \nonumber\]

or

\[5x^2+3x−2=(x+1)(5x−2) \nonumber\]

Analysis

This division problem had a remainder of 0. This tells us that the dividend is divided evenly by the divisor, and that the divisor is a factor of the dividend.

Example \(\PageIndex{2}\): Using Long Division to Divide a Third-Degree Polynomial

Divide \(6x^3+11x^2−31x+15\) by \(3x−2\).

Solution

\[\require{enclose}\begin{array} {rll}
\large2x^2 +\phantom{0}5x-\phantom{0}7 &\qquad & \large 6x^3\text{ divided by }3x\text{ is }2x^2.\\[-3pt]
\large 3x-2\enclose{longdiv}{ 6x^3+11x^2-31x+15} &\qquad &\\[-3pt]
\large \underline{-\left( 6x^3 -4x^2 \right)}\hspace{5.8em}&\qquad & \large\text{Multiply }3x-2\text{ by }2x^2. \\[-3pt]
\large 15x^2-31x \hspace{3.0em}&\qquad & \large\text{Subtract. Bring down next term. }15x^2\text{ divided by }3x\text{ is }5x. \\[-3pt]
\large \underline{-\left( 15x^2-10x\right)}\hspace{2.5em}&\qquad &\large\text{Multiply }3x-2\text{ by }5x. \\[-3pt]
\large -21x+15\hspace{0.5em}&\qquad & \large\text{Subtract. Bring down next term. }-21x\text{ divided by }3x\text{ is }-7. \\[-3pt]
\large \underline{-\left( -21x+14\right)} \hspace{0.1em}&\qquad & \large\text{Multiply }3x-2\text{by}-7.\\[-3pt]
\large 1\hspace{0.5em}&\qquad & \large\text{Subtract.The remainder is 1.}\\[8pt]
\end{array} \nonumber\]

There is a remainder of 1. We can express the result as:

\[\dfrac{6x^3+11x^2−31x+15}{3x−2}=2x^2+5x−7+\dfrac{1}{3x−2} \nonumber\]

Analysis

We can check our work by using the Division Algorithm to rewrite the solution. Then multiply.

\[(3x−2)(2x^2+5x−7)+1=6x^3+11x^2−31x+15 \nonumber\]

Notice, as we write our result,

• the dividend is \(6x^3+11x^2−31x+15\)
• the divisor is \(3x−2\)
• the quotient is \(2x^2+5x−7\)
• the remainder is \(1\)

Try It! \(\PageIndex{2}\)

Divide \(16x^3−12x^2+20x−3\) by \(4x+5\).

Solution

\(4x^2−8x+15−\dfrac{78}{4x+5}\)

Synthetic Division

Synthetic division is a shortcut that can be used when the divisor is a binomial in the form \(x−k\). In synthetic division, only the coefficients are used in the division process.

Given two polynomials, use synthetic division to divide

1. Write \(k\) for the divisor.
2. Write the coefficients of the dividend.
3. Bring the lead coefficient down.
4. Multiply the lead coefficient by \(k\). Write the product in the next column.
5. Add the terms of the second column.
6. Multiply the result by \(k\). Write the product in the next column.
7. Repeat steps 5 and 6 for the remaining columns.
8. 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 1, the next number from the right has degree 2, and so on.

Example \(\PageIndex{3}\): Using Synthetic Division to Divide a Second-Degree Polynomial

Use synthetic division to divide \(5x^2−3x−36\) by \(x−3\).

Solution

Begin by setting up the synthetic division. Write \(k\) and the coefficients.

Bring down the lead coefficient. Multiply the lead coefficient by \(k\).

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.

The result is \(5x+12\). The remainder is 0. So \(x−3\) is a factor of the original polynomial.

Analysis

Just as with long division, we can check our work by multiplying the quotient by the divisor and adding the remainder.

\[(x−3)(5x+12)+0=5x^2−3x−36 \nonumber\]

Example \(\PageIndex{4}\): Using Synthetic Division to Divide a Third-Degree Polynomial

Use synthetic division to divide \(4x^3+10x^2−6x−20\) 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.

\[ \large{\begin{array}{c}-2\\ \\ \\
\end{array}}
{\begin{align*}&\\[0pt]
&{\begin{array}{r|}\\[0pt]
\\[0pt] \end{array}}\\[1pt]
& \\[2pt]& \end{align*}}\!\!
{\begin{array}{rrrr}
1 &-1&-11&18\\
& 2&2&-18\\
\hline1&1 &-9&0
\end{array}}
\nonumber\]

The result is \(4x^2+2x−10\).

The remainder is 0. Thus, \(x+2\) is a factor of \(4x^3+10x^2−6x−20\).

Analysis

The graph of the polynomial function \(f(x)=4x^3+10x^2−6x−20\) in Figure \(\PageIndex{2}\) shows a zero at \(x=k=−2\). This confirms that \(x+2\) is a factor of \(4x^3+10x^2−6x−20\).

Example \(\PageIndex{5}\): Using Synthetic Division to Divide a Fourth-Degree Polynomial

Use synthetic division to divide \(−9x^4+10x^3+7x^2−6\) by \(x−1\).

Solution

Notice there is no \(x\)-term. We will use a zero as the coefficient for that term.

\[ \large{\begin{array}{c}1\\ \\ \\
\end{array}}
{\begin{align*}&\\[0pt]
&{\begin{array}{r|}\\[0pt]
\\[0pt] \end{array}}\\[1pt]
& \\[2pt]& \end{align*}}\!\!
{\begin{array}{rrrrr}
-9 & 10 &7&0 &-6\\
& -9&1&8 & 8\\
\hline-9&1&8& 8 &2
\end{array}}
\nonumber\]

The result is \(−9x^3+x^2+8x+8+\dfrac{2}{x−1}\).

Try It! \(\PageIndex{3}\)

Use synthetic division to divide \(3x^4+18x^3−3x+40\) by \(x+7\).

Solution

\(3x^3−3x^2+21x−150+\dfrac{1090}{x+7}\)

Example \(\PageIndex{6}\): Using Polynomial Division in an Application Problem

The volume of a rectangular solid is given by the polynomial \(3x^4−3x^3−33x^2+54x\). The length of the solid is given by \(3x\) and the width is given by \(x−2\). Find the height \(h\) of the solid.

Solution

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 \(\PageIndex{3}\). Let \(h\) equal the height of the box.

We can now write an equation by substituting the known values into the formula for the volume of a rectangular solid.

\[\begin{align*} V&=l{\cdot}w{\cdot}h \\ 3x^4−3x^3−33x^2+54x&=3x{\cdot}(x−2){\cdot}h \end{align*}\]

To solve for \(h\), first divide both sides by \(3x\).

\[\dfrac{3x{\cdot}(x−2){\cdot}h}{3x}=\dfrac{3x^4−3x^3−33x^2+54x}{3x} \nonumber\]

\[(x-2)h=\dfrac{x^3-x^2-11x+18}{x-2} \nonumber\]

Now solve for \(h\) using synthetic division.

\[h=\dfrac{x^3−x^2−11x+18}{x−2} \nonumber\]

\[ \large{\begin{array}{c}-2\\ \\ \\
\end{array}}
{\begin{align*}&\\[0pt]
&{\begin{array}{r|}\\[0pt]
\\[0pt] \end{array}}\\[1pt]
& \\[2pt]& \end{align*}}\!\!
{\begin{array}{rrrr}
1 &-1&-11&18\\
& 2&2&-18\\
\hline1&1 &-9&0
\end{array}}
\nonumber\]

The quotient is \(x^2+x−9\) and the remainder is \(0.\) The height of the solid is \(x^2+x−9\).

Try It! \(\PageIndex{4}\)

The area of a rectangle is given by \(3x^3+14x^2−23x+6\). The width of the rectangle is given by \(x+6\). Find an expression for the length of the rectangle.

Solution

\(3x^2−4x+1\)