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# Division of Polynomials

## Using long division to divide polynomials

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Practice Division of Polynomials
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Division of Polynomials

What if you had a polynomial like $2x^2 + 5x - 3$ and you wanted to divide it by a monomial like $x$ or a binomial like $x + 1$ ? How would you do so? After completing this Concept, you'll be able to divide polynomials like this one by monomials and binomials.

### Try This

To check your answers to long division problems involving polynomials, try the solver at http://calc101.com/webMathematica/long-divide.jsp . It shows the long division steps so you can tell where you may have made a mistake.

### Guidance

A rational expression is formed by taking the quotient of two polynomials.

Some examples of rational expressions are

$\frac{2x}{x^2-1} \qquad \frac{4x^2-3x+4}{2x} \qquad \frac{9x^2+4x-5}{x^2+5x-1} \qquad \frac{2x^3}{2x+3}$

Just as with rational numbers, the expression on the top is called the numerator and the expression on the bottom is called the denominator . In special cases we can simplify a rational expression by dividing the numerator by the denominator.

Divide a Polynomial by a Monomial

We’ll start by dividing a polynomial by a monomial. To do this, we divide each term of the polynomial by the monomial. When the numerator has more than one term, the monomial on the bottom of the fraction serves as the common denominator to all the terms in the numerator.

#### Example A

Divide.

a) $\frac{8x^2-4x+16}{2}$

b) $\frac{3x^2+6x-1}{x}$

c) $\frac{-3x^2-18x+6}{9x}$

Solution

a) $\frac{8x^2-4x+16}{2}=\frac{8x^2}{2}-\frac{4x}{2}+\frac{16}{2}=4x^2-2x+8$

b) $\frac{3x^3+6x-1}{x} = \frac{3x^3}{x}+\frac{6x}{x}-\frac{1}{x}=3x^2+6-\frac{1}{x}$

c) $\frac{-3x^2-18x+6}{9x}=-\frac{3x^2}{9x}-\frac{18x}{9x}+\frac{6}{9x}=-\frac{x}{3}-2+\frac{2}{3x}$

A common error is to cancel the denominator with just one term in the numerator.

Consider the quotient $\frac{3x+4}{4}$ .

Remember that the denominator of 4 is common to both the terms in the numerator. In other words we are dividing both of the terms in the numerator by the number 4.

The correct way to simplify is:

$\frac{3x+4}{4}=\frac{3x}{4}+\frac{4}{4}=\frac{3x}{4}+1$

A common mistake is to cross out the number 4 from the numerator and the denominator, leaving just $3x$ . This is incorrect, because the entire numerator needs to be divided by 4.

#### Example B

Divide $\frac{5x^3-10x^2+x-25}{-5x^2}$ .

Solution

$\frac{5x^3-10x^2+x-25}{-5x^2}=\frac{5x^3}{-5x^2}-\frac{10x^2}{-5x^2}+\frac{x}{-5x^2}-\frac{25}{-5x^2}$

The negative sign in the denominator changes all the signs of the fractions:

$-\frac{5x^3}{5x^2}+\frac{10x^2}{5x^2}-\frac{x}{5x^2}+\frac{25}{5x^2}=-x+2-\frac{1}{5x}+\frac{5}{x^2}$

Divide a Polynomial by a Binomial

We divide polynomials using a method that’s a lot like long division with numbers. We’ll explain the method by doing an example.

#### Example C

Divide $\frac{x^2+4x+5}{x+3}$ .

Solution

When we perform division, the expression in the numerator is called the dividend and the expression in the denominator is called the divisor .

To start the division we rewrite the problem in the following form:

$& {x+3 \overline{ ) x^2+4x+5 }}$

We start by dividing the first term in the dividend by the first term in the divisor: $\frac{x^2}{x}=x$ .

We place the answer on the line above the $x$ term:

$& \overset{\qquad x}{x+3 \overline{ ) x^2+4x+5 \;}}$

Next, we multiply the $x$ term in the answer by the divisor, $x + 3$ , and place the result under the dividend, matching like terms. $x$ times $(x + 3)$ is $x^2+3x$ , so we put that under the divisor:

$& \overset{\qquad x}{x+3 \overline{ ) x^2+4x+5 \;}}\\& \qquad \ \ x^2 + 3x$

Now we subtract $x^2+3x$ from $x^2+4x+5$ . It is useful to change the signs of the terms of $x^2+3x$ to $-x^2-3x$ and add like terms vertically:

$& \overset{\qquad x}{x+3 \overline{ ) x^2+4x+5 \;}}\\& \qquad \underline{-x^2 - 3x}\\& \qquad \qquad \quad \ x$

Now, we bring down the 5, the next term in the dividend.

$& \overset{\qquad x}{x+3 \overline{ ) x^2+4x+5 \;}}\\& \qquad \underline{-x^2 - 3x\;\;\;\;\;\;}\\& \qquad \qquad \quad \ x + 5$

And now we go through that procedure once more. First we divide the first term of $x + 5$ by the first term of the divisor. $x$ divided by $x$ is 1, so we place this answer on the line above the constant term of the dividend:

$& \overset{\qquad \qquad \quad x \ + \ 1}{x+3 \overline{ ) x^2+4x+5 \;}}\\& \qquad \underline{-x^2 - 3x\;\;\;\;\;\;}\\& \qquad \qquad \quad \ x + 5$

Multiply 1 by the divisor, $x + 3$ , and write the answer below $x + 5$ , matching like terms.

$& \overset{\qquad \qquad \quad x \ + \ 1}{x+3 \overline{ ) x^2+4x+5 \;}}\\& \qquad \underline{-x^2 - 3x\;\;\;\;\;\;}\\& \qquad \qquad \quad \ x + 5\\& \qquad \qquad \quad \ x + 3$

Subtract $x + 3$ from $x + 5$ by changing the signs of $x + 3$ to $-x -3$ and adding like terms:

$& \overset{\qquad \qquad \quad x \ + \ 1}{x+3 \overline{ ) x^2+4x+5 \;}}\\& \qquad \underline{-x^2 - 3x\;\;\;\;\;\;}\\& \qquad \qquad \quad \ x + 5\\& \qquad \qquad \ \ \underline{-x - 3}\\& \qquad \qquad \qquad \quad 2$

Since there are no more terms from the dividend to bring down, we are done. The quotient is $x + 1$ and the remainder is 2.

Remember that for a division with a remainder the answer is $\text{quotient}+\frac{\text{remainder}}{\text{divisor}}$ . So the answer to this division problem is $\frac{x^2+4x+5}{x+3}=x+1+\frac{2}{x+3}$ .

Check

To check the answer to a long division problem we use the fact that

$(\text{divisor} \times \text{quotient}) + \text{remainder} = \text{dividend}$

For the problem above, here’s how we apply that fact to check our solution:

$(x+3)(x+1)+2 & = x^2+4x+3+2\\& = x^2+4x+5$

Watch this video for help with the Examples above.

### Vocabulary

• A rational expression is formed by taking the quotient of two polynomials.

### Guided Practice

Divide $\frac{x^2+8x+17}{x+4}$ .

Solution

When we perform division, the expression in the numerator is called the dividend and the expression in the denominator is called the divisor .

To start the division we rewrite the problem in the following form:

$& {x+4 \overline{ ) x^2+8x+17 }}$

We start by dividing the first term in the dividend by the first term in the divisor: $\frac{x^2}{x}=x$ .

We place the answer on the line above the $x$ term:

$& \overset{\qquad x}{x+4 \overline{ ) x^2+8x+17 \;}}$

Next, we multiply the $x$ term in the answer by the divisor, $x + 4$ , and place the result under the dividend, matching like terms. $x$ times $(x + 4)$ is $x^2+4x$ , so we put that under the divisor:

$& \overset{\qquad x}{x+4 \overline{ ) x^2+8x+17\;}}\\& \qquad \ \ x^2 + 4x$

Now we subtract $x^2+4x$ from $x^2+8x+17$ . It is useful to change the signs of the terms of $x^2+4x$ to $-x^2-4x$ and add like terms vertically:

$& \overset{\qquad x}{x+4 \overline{ ) x^2+8x+17 \;}}\\& \qquad \underline{-x^2 - 4x}\\& \qquad \qquad \quad \ 4x$

Now, we bring down the 17, the next term in the dividend.

$& \overset{\qquad x}{x+4 \overline{ ) x^2+8x+17 \;}}\\& \qquad \underline{-x^2 - 4x\;\;\;\;\;\;}\\& \qquad \qquad \quad \ 4x + 17$

And now we go through that procedure once more. First we divide the first term of $4x + 17$ by the first term of the divisor. $4x$ divided by $x$ is 4, so we place this answer on the line above the constant term of the dividend:

$& \overset{\qquad \qquad \quad x \ + \ 4}{x+4 \overline{ ) x^2+8x+17 \;}}\\& \qquad \underline{-x^2 - 4x\;\;\;\;\;\;}\\& \qquad \qquad \quad \ x + 17$

Multiply 4 by the divisor, $x + 4$ , and write the answer below $4x + 16$ , matching like terms.

$& \overset{\qquad \qquad \quad x \ + \ 4}{x+4 \overline{ ) x^2+8x+17 \;}}\\& \qquad \underline{-x^2 - 4x\;\;\;\;\;\;}\\& \qquad \qquad \quad \ 4x + 17\\& \qquad \qquad \quad \ 4x + 16$

Subtract $4x + 16$ from $4x + 17$ by changing the signs of $4x + 16$ to $-4x -16$ and adding like terms:

$& \overset{\qquad \qquad \quad x \ + \ 4}{x+4 \overline{ ) x^2+8x+17 \;}}\\& \qquad \underline{-x^2 - 4x\;\;\;\;\;\;}\\& \qquad \qquad \quad \ x + 17\\& \qquad \qquad \ \ \underline{-4x - 16}\\& \qquad \qquad \qquad \quad 1$

Since there are no more terms from the dividend to bring down, we are done. The quotient is $x + 4$ and the remainder is 1.

Remember that for a division with a remainder the answer is $\text{quotient}+\frac{\text{remainder}}{\text{divisor}}$ . So the answer to this division problem is $\frac{x^2+8x+17}{x+4}=x+4+\frac{1}{x+4}$ .

Check

To check the answer to a long division problem we use the fact that

$(\text{divisor} \times \text{quotient}) + \text{remainder} = \text{dividend}$

For the problem above, here’s how we apply that fact to check our solution:

$(x+4)(x+4)+1 & = x^2+8x+16+1\\& = x^2+8x+17$

### Explore More

Divide the following polynomials:

1. $\frac{2x+4}{2}$
2. $\frac{x-4}{x}$
3. $\frac{5x-35}{5x}$
4. $\frac{x^2+2x-5}{x}$
5. $\frac{4x^2+12x-36}{-4x}$
6. $\frac{2x^2+10x+7}{2x^2}$
7. $\frac{x^3-x}{-2x^2}$
8. $\frac{5x^4-9}{3x}$
9. $\frac{x^3-12x^2+3x-4}{12x^2}$
10. $\frac{3-6x+x^3}{-9x^3}$
11. $\frac{x^2+3x+6}{x+1}$
12. $\frac{x^2-9x+6}{x-1}$
13. $\frac{x^2+5x+4}{x+4}$
14. $\frac{x^2-10x+25}{x-5}$
15. $\frac{x^2-20x+12}{x-3}$
16. $\frac{3x^2-x+5}{x-2}$
17. $\frac{9x^2+2x-8}{x+4}$
18. $\frac{3x^2-4}{3x+1}$
19. $\frac{5x^2+2x-9}{2x-1}$
20. $\frac{x^2-6x-12}{5x^4}$

### Vocabulary Language: English

Denominator

Denominator

The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. $\frac{5}{8}$ has denominator $8$.
Dividend

Dividend

In a division problem, the dividend is the number or expression that is being divided.
divisor

divisor

In a division problem, the divisor is the number or expression that is being divided into the dividend. For example: In the expression $152 \div 6$, 6 is the divisor and 152 is the dividend.
Polynomial long division

Polynomial long division

Polynomial long division is the standard method of long division, applied to the division of polynomials.
Rational Expression

Rational Expression

A rational expression is a fraction with polynomials in the numerator and the denominator.
Rational Root Theorem

Rational Root Theorem

The rational root theorem states that for a polynomial, $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, where $a_n, a_{n-1}, \cdots a_0$ are integers, the rational roots can be determined from the factors of $a_n$ and $a_0$. More specifically, if $p$ is a factor of $a_0$ and $q$ is a factor of $a_n$, then all the rational factors will have the form $\pm \frac{p}{q}$.
Remainder Theorem

Remainder Theorem

The remainder theorem states that if $f(k) = r$, then $r$ is the remainder when dividing $f(x)$ by $(x - k)$.
Synthetic Division

Synthetic Division

Synthetic division is a shorthand version of polynomial long division where only the coefficients of the polynomial are used.