Free Printable Polynomial Long Division Worksheets - Free Printable
Educational worksheet: Free Printable Polynomial Long Division Worksheets. Download and print for classroom or home learning activities.
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Step-by-step solution for: Free Printable Polynomial Long Division Worksheets
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Show Answer Key & Explanations
Step-by-step solution for: Free Printable Polynomial Long Division Worksheets
To solve the problems on this worksheet, we will use both long division and synthetic division where applicable. Let's go through each problem step by step.
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#### Solution:
We can use synthetic division since the divisor is of the form \( r + 5 \).
1. Write the coefficients of the dividend \( r^2 + 6r + 15 \): \( [1, 6, 15] \).
2. The root corresponding to \( r + 5 = 0 \) is \( r = -5 \).
3. Set up the synthetic division:
\[
\begin{array}{r|rrr}
-5 & 1 & 6 & 15 \\
& & -5 & -5 \\
\hline
& 1 & 1 & 10 \\
\end{array}
\]
- Bring down the first coefficient: \( 1 \).
- Multiply \( -5 \times 1 = -5 \) and add to the next coefficient: \( 6 + (-5) = 1 \).
- Multiply \( -5 \times 1 = -5 \) and add to the next coefficient: \( 15 + (-5) = 10 \).
The quotient is \( r + 1 \) and the remainder is \( 10 \).
Thus, the result is:
\[
(r^2 + 6r + 15) \div (r + 5) = r + 1 + \frac{10}{r + 5}
\]
---
#### Solution:
Again, use synthetic division with the root \( r = -7 \).
1. Coefficients of the dividend \( r^2 + 10r + 13 \): \( [1, 10, 13] \).
2. Set up the synthetic division:
\[
\begin{array}{r|rrr}
-7 & 1 & 10 & 13 \\
& & -7 & -21 \\
\hline
& 1 & 3 & -8 \\
\end{array}
\]
- Bring down the first coefficient: \( 1 \).
- Multiply \( -7 \times 1 = -7 \) and add to the next coefficient: \( 10 + (-7) = 3 \).
- Multiply \( -7 \times 3 = -21 \) and add to the next coefficient: \( 13 + (-21) = -8 \).
The quotient is \( r + 3 \) and the remainder is \( -8 \).
Thus, the result is:
\[
(r^2 + 10r + 13) \div (r + 7) = r + 3 - \frac{8}{r + 7}
\]
---
#### Solution:
Use synthetic division with the root \( n = 9 \).
1. Coefficients of the dividend \( n^3 - 5n^2 - 33n - 37 \): \( [1, -5, -33, -37] \).
2. Set up the synthetic division:
\[
\begin{array}{r|rrrr}
9 & 1 & -5 & -33 & -37 \\
& & 9 & 36 & 27 \\
\hline
& 1 & 4 & 3 & -10 \\
\end{array}
\]
- Bring down the first coefficient: \( 1 \).
- Multiply \( 9 \times 1 = 9 \) and add to the next coefficient: \( -5 + 9 = 4 \).
- Multiply \( 9 \times 4 = 36 \) and add to the next coefficient: \( -33 + 36 = 3 \).
- Multiply \( 9 \times 3 = 27 \) and add to the next coefficient: \( -37 + 27 = -10 \).
The quotient is \( n^2 + 4n + 3 \) and the remainder is \( -10 \).
Thus, the result is:
\[
(n^3 - 5n^2 - 33n - 37) \div (n - 9) = n^2 + 4n + 3 - \frac{10}{n - 9}
\]
---
#### Solution:
Use synthetic division with the root \( x = -10 \).
1. Coefficients of the dividend \( x^3 + 6x^2 - 30x + 102 \): \( [1, 6, -30, 102] \).
2. Set up the synthetic division:
\[
\begin{array}{r|rrrr}
-10 & 1 & 6 & -30 & 102 \\
& & -10 & 40 & -100 \\
\hline
& 1 & -4 & 10 & 2 \\
\end{array}
\]
- Bring down the first coefficient: \( 1 \).
- Multiply \( -10 \times 1 = -10 \) and add to the next coefficient: \( 6 + (-10) = -4 \).
- Multiply \( -10 \times (-4) = 40 \) and add to the next coefficient: \( -30 + 40 = 10 \).
- Multiply \( -10 \times 10 = -100 \) and add to the next coefficient: \( 102 + (-100) = 2 \).
The quotient is \( x^2 - 4x + 10 \) and the remainder is \( 2 \).
Thus, the result is:
\[
(x^3 + 6x^2 - 30x + 102) \div (x + 10) = x^2 - 4x + 10 + \frac{2}{x + 10}
\]
---
\[
\boxed{
\begin{aligned}
1. & \quad r + 1 + \frac{10}{r + 5} \\
2. & \quad r + 3 - \frac{8}{r + 7} \\
3. & \quad n^2 + 4n + 3 - \frac{10}{n - 9} \\
4. & \quad x^2 - 4x + 10 + \frac{2}{x + 10}
\end{aligned}
}
\]
---
If you need solutions for the remaining problems, please let me know!
---
Problem 1: \( (r^2 + 6r + 15) \div (r + 5) \)
#### Solution:
We can use synthetic division since the divisor is of the form \( r + 5 \).
1. Write the coefficients of the dividend \( r^2 + 6r + 15 \): \( [1, 6, 15] \).
2. The root corresponding to \( r + 5 = 0 \) is \( r = -5 \).
3. Set up the synthetic division:
\[
\begin{array}{r|rrr}
-5 & 1 & 6 & 15 \\
& & -5 & -5 \\
\hline
& 1 & 1 & 10 \\
\end{array}
\]
- Bring down the first coefficient: \( 1 \).
- Multiply \( -5 \times 1 = -5 \) and add to the next coefficient: \( 6 + (-5) = 1 \).
- Multiply \( -5 \times 1 = -5 \) and add to the next coefficient: \( 15 + (-5) = 10 \).
The quotient is \( r + 1 \) and the remainder is \( 10 \).
Thus, the result is:
\[
(r^2 + 6r + 15) \div (r + 5) = r + 1 + \frac{10}{r + 5}
\]
---
Problem 2: \( (r^2 + 10r + 13) \div (r + 7) \)
#### Solution:
Again, use synthetic division with the root \( r = -7 \).
1. Coefficients of the dividend \( r^2 + 10r + 13 \): \( [1, 10, 13] \).
2. Set up the synthetic division:
\[
\begin{array}{r|rrr}
-7 & 1 & 10 & 13 \\
& & -7 & -21 \\
\hline
& 1 & 3 & -8 \\
\end{array}
\]
- Bring down the first coefficient: \( 1 \).
- Multiply \( -7 \times 1 = -7 \) and add to the next coefficient: \( 10 + (-7) = 3 \).
- Multiply \( -7 \times 3 = -21 \) and add to the next coefficient: \( 13 + (-21) = -8 \).
The quotient is \( r + 3 \) and the remainder is \( -8 \).
Thus, the result is:
\[
(r^2 + 10r + 13) \div (r + 7) = r + 3 - \frac{8}{r + 7}
\]
---
Problem 3: \( (n^3 - 5n^2 - 33n - 37) \div (n - 9) \)
#### Solution:
Use synthetic division with the root \( n = 9 \).
1. Coefficients of the dividend \( n^3 - 5n^2 - 33n - 37 \): \( [1, -5, -33, -37] \).
2. Set up the synthetic division:
\[
\begin{array}{r|rrrr}
9 & 1 & -5 & -33 & -37 \\
& & 9 & 36 & 27 \\
\hline
& 1 & 4 & 3 & -10 \\
\end{array}
\]
- Bring down the first coefficient: \( 1 \).
- Multiply \( 9 \times 1 = 9 \) and add to the next coefficient: \( -5 + 9 = 4 \).
- Multiply \( 9 \times 4 = 36 \) and add to the next coefficient: \( -33 + 36 = 3 \).
- Multiply \( 9 \times 3 = 27 \) and add to the next coefficient: \( -37 + 27 = -10 \).
The quotient is \( n^2 + 4n + 3 \) and the remainder is \( -10 \).
Thus, the result is:
\[
(n^3 - 5n^2 - 33n - 37) \div (n - 9) = n^2 + 4n + 3 - \frac{10}{n - 9}
\]
---
Problem 4: \( (x^3 + 6x^2 - 30x + 102) \div (x + 10) \)
#### Solution:
Use synthetic division with the root \( x = -10 \).
1. Coefficients of the dividend \( x^3 + 6x^2 - 30x + 102 \): \( [1, 6, -30, 102] \).
2. Set up the synthetic division:
\[
\begin{array}{r|rrrr}
-10 & 1 & 6 & -30 & 102 \\
& & -10 & 40 & -100 \\
\hline
& 1 & -4 & 10 & 2 \\
\end{array}
\]
- Bring down the first coefficient: \( 1 \).
- Multiply \( -10 \times 1 = -10 \) and add to the next coefficient: \( 6 + (-10) = -4 \).
- Multiply \( -10 \times (-4) = 40 \) and add to the next coefficient: \( -30 + 40 = 10 \).
- Multiply \( -10 \times 10 = -100 \) and add to the next coefficient: \( 102 + (-100) = 2 \).
The quotient is \( x^2 - 4x + 10 \) and the remainder is \( 2 \).
Thus, the result is:
\[
(x^3 + 6x^2 - 30x + 102) \div (x + 10) = x^2 - 4x + 10 + \frac{2}{x + 10}
\]
---
Final Answer for Problems 1-4:
\[
\boxed{
\begin{aligned}
1. & \quad r + 1 + \frac{10}{r + 5} \\
2. & \quad r + 3 - \frac{8}{r + 7} \\
3. & \quad n^2 + 4n + 3 - \frac{10}{n - 9} \\
4. & \quad x^2 - 4x + 10 + \frac{2}{x + 10}
\end{aligned}
}
\]
---
If you need solutions for the remaining problems, please let me know!
Parent Tip: Review the logic above to help your child master the concept of long division with polynomials worksheet.