Lesson 5.4 : Dividing Polynomials

New Concept

Mastering polynomial division allows you to break down complex expressions into simpler factors. You will learn two key methods: traditional long division and the efficient synthetic division, both based on the Division Algorithm, $f(x) = d(x)q(x) + r(x)$.

What’s next

Now that you have the basics, you'll work through interactive examples of long division and then master synthetic division with a series of practice cards.

Property

Division of polynomials that contain more than one term has similarities to long division of whole numbers. We can write a polynomial dividend as the product of the divisor and the quotient added to the remainder. To divide a polynomial by a binomial using long division:

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.

Examples

• To divide $3x^2 + 7x - 6$ by $x + 3$, the long division process yields a quotient of $3x - 2$ and a remainder of $0$.

• To divide $4x^3 - 5x^2 + 8x - 10$ by $x - 1$, the process results in a quotient of $4x^2 - x + 7$ and a remainder of $-3$. The final answer is written as $4x^2 - x + 7 - \frac{3}{x-1}$.

Property

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

$q(x)$ 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 $f(x)$.

Examples

• Dividing $f(x) = x^2 - 9$ by $d(x) = x-3$ gives a quotient $q(x) = x+3$ and remainder $r(x) = 0$. The relationship is $x^2 - 9 = (x-3)(x+3) + 0$.

• Dividing $f(x) = 2x^2 + 5x + 7$ by $d(x) = x+2$ gives a quotient $q(x) = 2x+1$ and remainder $r(x) = 5$. This is expressed as $2x^2+5x+7 = (x+2)(2x+1) + 5$.

Property

Synthetic division is a shortcut that can be used when the divisor is a binomial in the form $x - k$ where $k$ is a real number. In synthetic division, only the coefficients are used in the division process.

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.

Examples

• To divide $2x^2 - 5x - 3$ by $x - 3$, we use $k=3$ in synthetic division. The bottom row of numbers represents the quotient $2x+1$ and a remainder of $0$.

• To divide $x^3 + 8x^2 - 5x + 10$ by $x + 2$, we use $k=-2$. The process yields a quotient of $x^2+6x-17$ and a remainder of $44$. The result is $x^2+6x-17 + \frac{44}{x+2}$.

Property

Polynomial division can be used to solve a variety of application problems involving expressions for area and volume. For example, if the volume of a rectangular solid is given by the polynomial $V(x)$ and its length and width are given by polynomials $l(x)$ and $w(x)$, the height $h(x)$ can be found by dividing the volume by the product of the length and width.

This often requires dividing the volume by one dimension, and then dividing the resulting quotient by the second dimension.

Examples

• The area of a rectangle is $A(x) = 3x^2 + 8x - 3$ and its length is $l(x) = x+3$. The width is $w(x) = A(x) \div l(x) = (3x^2 + 8x - 3) \div (x+3) = 3x-1$.

• The volume of a box is $V(x) = 2x^3 + 7x^2 + 2x - 3$. Its length is $x+1$ and width is $x+3$. The height is $h(x) = V(x) \div ((x+1)(x+3)) = (2x^3 + 7x^2 + 2x - 3) \div (x^2+4x+3) = 2x-1$.