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Algebra 1

6.3.2 Dividing Polynomials Using Long Division

Algebra 16.3.2 Dividing Polynomials Using Long Division
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6.3.2 • Dividing Polynomials Using Long Division

Activity

Recall the process of dividing 960 ÷ 20 960 ÷ 20 as a class using long division. Use this same method to divide the polynomials below. After each problem, check in with a partner. With your partner, discuss your work and use the words dividend, divisor, and quotient in your explanations.

1. Find the quotient using long division.

( m 2 + 9 m + 20 ) ÷ ( m + 4 ) ( m 2 + 9 m + 20 ) ÷ ( m + 4 )

Look through the video for help.

2. Find the quotient using long division.

( p 2 + 2 p 8 ) ÷ ( p 2 7 p + 10 ) ( p 2 + 2 p 8 ) ÷ ( p 2 7 p + 10 )

3. Find the quotient using long division.

( 5 x + 5 ) ÷ ( x + 5 ) ( 5 x + 5 ) ÷ ( x + 5 )

Notice in the next problem, there is no x 3 x 3 term in the dividend. Add 0 x 3 0 x 3 as a placeholder when using long division.

4. Find the quotient using long division.

( x 4 11 x 2 7 x 6 ) ÷ ( x + 3 ) ( x 4 11 x 2 7 x 6 ) ÷ ( x + 3 )

In this next problem, you will need to add two placeholders to divide.

5. Find the quotient using long division.

( 125 x 3 8 ) ÷ ( 5 x 2 ) ( 125 x 3 8 ) ÷ ( 5 x 2 )

Are you ready for more?

Extending Your Thinking

Find the quotient using long division.

( 125 y 5 80 y 4 31 y 3 + 37 y 2 + 10 y 24 ) ÷ ( 5 y 4 ) ( 125 y 5 80 y 4 31 y 3 + 37 y 2 + 10 y 24 ) ÷ ( 5 y 4 )

Video: Dividing Polynomials Using Long Division

Watch the following video to learn more about how to divide polynomials using long division.

Self Check

Find the quotient: ( 3 x 3 19 x 2 9 x 35 ) ÷ ( x 7 ) .
  1. x 2 + 2 x + 5
  2. 3 x 2 + 2 x 23
  3. 3 x 2 + 2 x + 5
  4. 3 x 2 + 40 x + 5

Additional Resources

Dividing Polynomials Using Long Division

To divide a polynomial by a binomial, we follow a procedure very similar to long division of numbers. So let’s look carefully at the steps we take when we divide a 3 3 -digit number, 875 875 , by a 2 2 -digit number, 25 25 .

Step 1 - Starting from the leftmost digit in the dividend and moving right one digit at a time, find the smallest value possible that is larger than the divisor.

Divide this value, 87 87 , by the divisor, 25 25 . Write the whole-number result with no remainder in the space for the quotient.

Step 2 -Multiply this digit in the quotient by the divisor, 3 · 25 3 · 25 .

Write the result below the corresponding digits in the dividend.

Step 3 -Subtract the two values.

Step 4 - Bring down the next digit in the dividend.

Step 5 - Repeat the division by dividing the new value, 125 125 , by the divisor.

Write the result, 5 5 , over the corresponding digit in the quotient.

Step 6 - Multiply this digit in the quotient by the divisor, 5 · 25 5 · 25 .

Write the result at the bottom below the previous number.

Step 7 -Subtract the two values. Since there are no more digits to bring down, this value is the remainder.

The quotient is on top.

We check division by multiplying the quotient by the divisor.

If we did the division correctly, the product should equal the dividend.

  • 35 · 25 35 · 25
  • 875 875

Now we will divide a trinomial by a binomial. As you read through the Example, notice how similar the steps are to the numerical Example above.

Example 1

Find the quotient: ( x 2 + 9 x + 20 ) ÷ ( x + 5 ) ( x 2 + 9 x + 20 ) ÷ ( x + 5 ) .

Step 1 - Write it as a long division problem. Be sure the dividend is in standard form.

Step 2 - Divide x 2 x 2 by x x . It may help to ask yourself, “What do I need to multiply x x by to get x 2 x 2 ?”

Step 3 - Put the answer, x x , in the quotient over the x x term. Multiply x x times x + 5 x + 5 . Line up the like terms under the dividend.

Step 4 - Subtract x 2 + 5 x x 2 + 5 x from x 2 + 9 x x 2 + 9 x . You may find it easier to change the signs and then add. Then bring down the last term, 20 20 .

Step 5 - Divide 4 x 4 x by x x . It may help to ask yourself, “What do I need to multiply x x by to get 4 x 4 x ?” Put the answer, 4 4 , in the quotient over the constant term.

Step 6 - Multiply 4 4 times x + 5 x + 5 .

Step 7 - Subtract 4 x + 20 4 x + 20 from 4 x + 20 4 x + 20 .

Step 8 - Check: Multiply the quotient by the divisor. You should get the dividend.

( x + 4 ) ( x + 5 ) ( x + 4 ) ( x + 5 )

x 2 + 9 x + 20 x 2 + 9 x + 20

When we divided 875 875 by 25 25 , we had no remainder. But sometimes division of numbers does leave a remainder. The same is true when we divide polynomials. In the next Example, we’ll have division that leaves a remainder. We write the remainder as a fraction with the divisor as the denominator.

Look back at the dividend in the previous Example. The terms were written in descending order of degrees, and there were no missing degrees. The dividend in the next Example will be x 4 x 2 + 5 x 6 x 4 x 2 + 5 x 6 . It is missing an x 3 x 3 term. We will add in 0 x 3 0 x 3 as a placeholder.

Example 2

Find the quotient: ( x 4 x 2 + 5 x 6 ) ÷ ( x + 2 ) ( x 4 x 2 + 5 x 6 ) ÷ ( x + 2 ) .

Notice that there is no x 3 x 3 term in the dividend. We will add 0 x 3 0 x 3 as a placeholder.

Step 1 - Write it as a long division problem.

Be sure the dividend is in standard form with placeholders for missing terms.

Step 2 - Divide x 4 x 4 by x x .

Put the answer, x 3 x 3 , in the quotient over the x 3 x 3 term. Multiply x 3 x 3 times x + 2 x + 2 .

Line up the like terms. Subtract and then bring down the next term.

Step 3 - Divide 2 x 3 2 x 3 by x x .

Put the answer, 2 x 2 2 x 2 , in the quotient over the x 2 x 2 term.

Multiply 2 x 2 2 x 2 times x + 1 x + 1 . Line up the like terms.

Subtract and bring down the next term.

Step 4 - Divide 3 x 2 3 x 2 by x x .

Put the answer, 3 x 3 x , in the quotient over the x x term.

Multiply 3 x 3 x times x + 1 x + 1 .

Line up the like terms. Subtract and bring down the next term.

Step 5 - Divide x x by x x . Put the answer, 1 1 , in the quotient over the constant term.

Multiply 1 1 times x + 1 x + 1 . Line up the like terms.

Change the signs and then add.

Write the remainder as a fraction with the divisor as the denominator.

Step 6 - To check, multiply ( x + 2 ) ( x 3 2 x 2 + 3 x 1 4 x + 2 ) ( x + 2 ) ( x 3 2 x 2 + 3 x 1 4 x + 2 ) .

The result should be x 4 x 2 + 5 x 6 x 4 x 2 + 5 x 6 .

In the next Example, we will divide by 2 a + 3 2 a + 3 .

As we divide, we will have to consider the coefficients as well as the variables.

Example 3

Find the quotient: ( 8 a 3 + 27 ) ÷ ( 2 a + 3 ) ( 8 a 3 + 27 ) ÷ ( 2 a + 3 ) .

This time, we will show the division all in one step. We need to add two placeholders to divide.

To check, multiply ( 2 a + 3 ) ( 4 a 2 6 a + 9 ) ( 2 a + 3 ) ( 4 a 2 6 a + 9 ) .

The result should be 8 a 3 + 27 8 a 3 + 27 .

Example 4

Find the quotient: ( 2 x + 14 ) ÷ ( x + 7 ) ( 2 x + 14 ) ÷ ( x + 7 ) . Can you see a quick way to solve this?

\begin{array}{r} 2\phantom{)} \\ x+7{\overline{\smash{\big)}\,2x+14}}\\ \underline{-~\phantom{(}(2x+14)}\\ 0\\ \end{array}

The answer is 2.

Try it

Dividing Polynomials Using Long Division

Find the quotients.

1. ( y 2 + 10 y + 21 ) ÷ ( y + 3 ) ( y 2 + 10 y + 21 ) ÷ ( y + 3 )

2. ( x 4 7 x 2 + 7 x + 6 ) ÷ ( x + 3 ) ( x 4 7 x 2 + 7 x + 6 ) ÷ ( x + 3 )

3. ( x 3 64 ) ÷ ( x 4 ) ( x 3 64 ) ÷ ( x 4 )

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