Dividing Polynomial Functions

Table of Contents

You can divide a polynomial just like numbers and dividing a polynomial will give an expression as quotient and remainder. The polynomial you are going to divide must have more terms than the divisor, else the division will not be fruitful.

In this article, you will learn about polynomial long division and synthetic division techniques.

Polynomial Long Division

To divide a polynomial, you need to follow some steps.

Example #1

Step 1: Keep the terms of the dividend and divisor in standard form, that is, in the descending powers of the variable.

x + 2 )\overline {x^2 + 7x + 10}

Step 2: Take first term of the dividend and divide with the divisor, you will get the first term of the quotient. Do it as if you are dividing two numbers.

x\\ x + 2 )\overline {x^2 + 7x + 10}

Step 3: Multiply every term of the divisor with the first term of the quotient.

x\\ x + 2 )\overline{x^2 + 7x + 10}\\ x^2 + 2x

Step 4: Subtract the result of multiplication in step 3 with dividend.

x\\ x + 2 ) \overline{x^2 + 7x + 10} \\ x^2 + 2x\\ \hspace{1.5 cm}\overline{5x + 10}

Step 5: Bring down the new term from the dividend and repeat the step 1 to 5 until you get a remainder 0 or some other value.

x + 5\\ x + 2 ) \overline{x^2 + 7x + 10} \\ x^2 + 2x\\ \hspace{2 cm}\overline{5x + 10} \\ \hspace{2 cm}5x + 10\\ \hspace{3 cm}\overline{0}\\

The solution to the above polynomial is $x + 5$ which is you obtained after the polynomial long division as quotient. Let perform a polynomial long division on another polynomial.

Example #2:

Divide \hspace{3 mm} x - 1 ) \overline{x^3 + 3x^2 + 3x -4}

Solution:

\begin{aligned} &\hspace{ 1 cm}x^2 + 4x + 7 \\ &x - 1 ) \overline{x^3 + 3x^2 + 3x - 4} \\ &\hspace{ 1 cm}x^3 - x^2 \\ &\hspace{ 1 cm}\overline{4x^2 + 3x } \\ &\hspace{ 1 cm}4x^2 - 4x\\ &\hspace{ 2 cm}\overline{7x - 4} \\ &\hspace{ 2 cm}7x - 7\\ &\hspace{ 2.7 cm}\overline{11}\\ \end{aligned}

This time the solution is a trinomial $x^2 + 4x + 7$ . Note that the result of polynomial long division is not always zero, you may get a non-zero remainder too.

How do you write answer when the polynomial long division gives you a remainder. As in above case, you can write answers as

\frac{x^3 + 3x^2 + 3x -4}{x - 1} = x^2 + 4x + 7 + \frac{13}{x - 1}

The divisor still tries to divide the remainder so you represent it as a term.

Division Algorithm

We can rewrite the whole dividend , divisor , quotient and remainder as an expression by itself.

x^3 + 3x^2 + 3x -4 = (x - 1)(x^2 + 4x + 7) +11

What we are doing is a check to see whether multiplying and adding the remainder back will give us the original polynomial function $p(x)$

Let say that the dividend is $p(x)$ , divisor is $d(x)$ , quotient is $q(x)$ and remainder is $r(x)$ .

The degree of divisor $d(x)$ is less than or equal to $p(x)$ , where $d(x) \neq 0$ . Also, there exists unique polynomials $q(x)$ and $r(x)$ .

The degree of remainder $f(x)$ is 0 or less than degree of divisor $d(x)$ . If remainder $r(x) = 0$ , then we can say that divisor $d(x)$ divides polynomial $p(x)$ evenly and $q(x)$ and $d(x)$ are factors of the polynomial.

Synthetic Division

Another way to divide polynomial is much faster and efficient provide that the divisor is in the form $(x - c)$ where $c$ is a constant.

We can take our previous example to perform the synthetic division.

Example #3

Divide using synthetic division method: $x - 1 ) \overline{x^3 + 3x^2 + 3x -4}$

Solution:

Step 1: Write the constant $c$ from the divisor and all the coefficients from the dividend.

Divisor c value	1st term coefficient	2nd term coefficient	3rd term coefficient	4th term coefficient
1	1	3	3	4

Step 3: Write the 1st term coefficient in third column.

Divisor c value	1st term coefficient	2nd term coefficient	3rd term coefficient	4th term coefficient
1		3	3	4

	1

Step 4: Multiply 1st term coefficient with c and write the result in second column and second row and add the entries of second column. Write the result in second column, third row.

Divisor c value	1st term coefficient	2nd term coefficient	3rd term coefficient	4th term coefficient
1		3	3	4
		1
	1	4

Step 5: Repeat the process from step 1 to 4.

Divisor c value	1st term coefficient	2nd term coefficient	3rd term coefficient	4th term coefficient
1		3	3	4
		1	4	7
	1	4	7	11

The numbers in the last row is coefficients of quotient $q(x)$ and remainder $r(x)$ . Therefore, the result is

q(x) = x^2 + 4x + 7 + \frac{11}{x -1}

Remainder Theorem

Let us remember the division algorithm, which is

f(x) = d(x) . q(x) + r

Suppose we are dividing $f(x)$ with $x - c$ , then remainder will be a constant $r$ , which means,

f(x) = (x - c) q(x) + r

Given the above equation, suppose $x = c$ , then

f(c) = (c - c)q(c) + r\\ = 0 \cdot q(c) + r\\ f(x) = r

Therefore, remainder theorem states that you can use $c$ from $x - c$ and get the remainder.

Example #4

Find the remainder for $x^3 + 4x^2 + 3x + 2$ divided by $x - 1$

Solution:

Using reminder theorem, $c = 1$ and $f(x) = r$ .

f(1) = 1^3 + 4(1)^2 + 3(1) + 2 = 1 + 4 + 3 + 2 = 10

Therefore, remainder is $10$ .

Verify the results:

1 : 1  4  3  2
       1  5  8
    1  5  8  10

Using the synthetic division we find that the remainder is 10.

Factor Theorem

The factor theorem is derived from division algorithm. Suppose the $x - c$ divides the polynomial $f(x)$ . We know that by remainder theorem, $f(x) = (x - c) q(x) + r$ which result in $f(x) = r$ .

Let us replace $r$ in above equation as $f(c)$ . Now the division algorithm becomes $f(x) = (x -c ) q(x) + f(c)$ .

Suppose if the $f(c) = 0$ then, the equation becomes $f(x) = (x - c) q(x)$ which implies that $x-c$ is a factor of $f(x)$ .

Let us say that $x - c$ is a factor of polynomial $f(x)$ .Then,

f(x) = (x - c) q(x) , if \hspace{2 mm} x = c\\ f(c) = (c - c) q(c) = 0 \cdot q(c) = 0

If $x - c$ is a factor of $f(x)$ , then $f(c) = 0$ . This is known as factor theorem.

Summary

In this article, you learned about polynomial long division, synthetic division, division algorithm and two theorem that are derived from the division algorithms – remainder theorem and factor theorem.