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Synthetic Division Worksheets - Math Monks - Free Printable

Synthetic Division Worksheets - Math Monks

Educational worksheet: Synthetic Division Worksheets - Math Monks. Download and print for classroom or home learning activities.

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Show Answer Key & Explanations Step-by-step solution for: Synthetic Division Worksheets - Math Monks

Problem: Solve the given polynomial division problems using synthetic division.



#### Step-by-Step Solutions:

---

Problem 1:


\[
(x^3 - 2x^2 - 5x + 6) \div (x - 3)
\]

#### Synthetic Division Setup:
- Divisor: \( x - 3 \) → Root: \( r = 3 \)
- Coefficients of the dividend: \( 1, -2, -5, 6 \)

\[
\begin{array}{r|rrrr}
3 & 1 & -2 & -5 & 6 \\
& & 3 & 3 & -6 \\
\hline
& 1 & 1 & -2 & 0 \\
\end{array}
\]

#### Explanation:
1. Bring down the leading coefficient: \( 1 \).
2. Multiply \( 1 \times 3 = 3 \) and add to the next coefficient: \( -2 + 3 = 1 \).
3. Multiply \( 1 \times 3 = 3 \) and add to the next coefficient: \( -5 + 3 = -2 \).
4. Multiply \( -2 \times 3 = -6 \) and add to the next coefficient: \( 6 + (-6) = 0 \).

#### Result:
The quotient is \( x^2 + x - 2 \) and the remainder is \( 0 \).

\[
\boxed{x^2 + x - 2}
\]

---

Problem 2:


\[
(x^4 - 5x^3 + 7x^2 - 34x - 1) \div (x - 5)
\]

#### Synthetic Division Setup:
- Divisor: \( x - 5 \) → Root: \( r = 5 \)
- Coefficients of the dividend: \( 1, -5, 7, -34, -1 \)

\[
\begin{array}{r|rrrrr}
5 & 1 & -5 & 7 & -34 & -1 \\
& & 5 & 0 & 35 & 5 \\
\hline
& 1 & 0 & 7 & 1 & -6 \\
\end{array}
\]

#### Explanation:
1. Bring down the leading coefficient: \( 1 \).
2. Multiply \( 1 \times 5 = 5 \) and add to the next coefficient: \( -5 + 5 = 0 \).
3. Multiply \( 0 \times 5 = 0 \) and add to the next coefficient: \( 7 + 0 = 7 \).
4. Multiply \( 7 \times 5 = 35 \) and add to the next coefficient: \( -34 + 35 = 1 \).
5. Multiply \( 1 \times 5 = 5 \) and add to the next coefficient: \( -1 + 5 = 4 \).

#### Result:
The quotient is \( x^3 + 0x^2 + 7x + 1 \) (or simply \( x^3 + 7x + 1 \)) and the remainder is \( -6 \).

\[
\boxed{x^3 + 7x + 1 \text{ R } -6}
\]

---

Problem 3:


\[
(7x^3 + 4x + 8) \div (x + 2)
\]

#### Synthetic Division Setup:
- Divisor: \( x + 2 \) → Root: \( r = -2 \)
- Coefficients of the dividend: \( 7, 0, 4, 8 \) (Note: Include a placeholder for \( x^2 \)).

\[
\begin{array}{r|rrrr}
-2 & 7 & 0 & 4 & 8 \\
& & -14 & 28 & -64 \\
\hline
& 7 & -14 & 32 & -56 \\
\end{array}
\]

#### Explanation:
1. Bring down the leading coefficient: \( 7 \).
2. Multiply \( 7 \times (-2) = -14 \) and add to the next coefficient: \( 0 + (-14) = -14 \).
3. Multiply \( -14 \times (-2) = 28 \) and add to the next coefficient: \( 4 + 28 = 32 \).
4. Multiply \( 32 \times (-2) = -64 \) and add to the next coefficient: \( 8 + (-64) = -56 \).

#### Result:
The quotient is \( 7x^2 - 14x + 32 \) and the remainder is \( -56 \).

\[
\boxed{7x^2 - 14x + 32 \text{ R } -56}
\]

---

Problem 4:


\[
(x^3 + 5x^2 + 7x + 2) \div (x + 2)
\]

#### Synthetic Division Setup:
- Divisor: \( x + 2 \) → Root: \( r = -2 \)
- Coefficients of the dividend: \( 1, 5, 7, 2 \)

\[
\begin{array}{r|rrrr}
-2 & 1 & 5 & 7 & 2 \\
& & -2 & -6 & -2 \\
\hline
& 1 & 3 & 1 & 0 \\
\end{array}
\]

#### Explanation:
1. Bring down the leading coefficient: \( 1 \).
2. Multiply \( 1 \times (-2) = -2 \) and add to the next coefficient: \( 5 + (-2) = 3 \).
3. Multiply \( 3 \times (-2) = -6 \) and add to the next coefficient: \( 7 + (-6) = 1 \).
4. Multiply \( 1 \times (-2) = -2 \) and add to the next coefficient: \( 2 + (-2) = 0 \).

#### Result:
The quotient is \( x^2 + 3x + 1 \) and the remainder is \( 0 \).

\[
\boxed{x^2 + 3x + 1}
\]

---

Problem 5:


\[
(3x^2 + 7x - 20) \div (x + 5)
\]

#### Synthetic Division Setup:
- Divisor: \( x + 5 \) → Root: \( r = -5 \)
- Coefficients of the dividend: \( 3, 7, -20 \)

\[
\begin{array}{r|rrr}
-5 & 3 & 7 & -20 \\
& & -15 & 40 \\
\hline
& 3 & -8 & 20 \\
\end{array}
\]

#### Explanation:
1. Bring down the leading coefficient: \( 3 \).
2. Multiply \( 3 \times (-5) = -15 \) and add to the next coefficient: \( 7 + (-15) = -8 \).
3. Multiply \( -8 \times (-5) = 40 \) and add to the next coefficient: \( -20 + 40 = 20 \).

#### Result:
The quotient is \( 3x - 8 \) and the remainder is \( 20 \).

\[
\boxed{3x - 8 \text{ R } 20}
\]

---

Problem 6:


\[
(7x^3 + 6x - 8) \div (x - 4)
\]

#### Synthetic Division Setup:
- Divisor: \( x - 4 \) → Root: \( r = 4 \)
- Coefficients of the dividend: \( 7, 0, 6, -8 \) (Note: Include placeholders for \( x^2 \) and \( x^0 \)).

\[
\begin{array}{r|rrrr}
4 & 7 & 0 & 6 & -8 \\
& & 28 & 112 & 472 \\
\hline
& 7 & 28 & 118 & 464 \\
\end{array}
\]

#### Explanation:
1. Bring down the leading coefficient: \( 7 \).
2. Multiply \( 7 \times 4 = 28 \) and add to the next coefficient: \( 0 + 28 = 28 \).
3. Multiply \( 28 \times 4 = 112 \) and add to the next coefficient: \( 6 + 112 = 118 \).
4. Multiply \( 118 \times 4 = 472 \) and add to the next coefficient: \( -8 + 472 = 464 \).

#### Result:
The quotient is \( 7x^2 + 28x + 118 \) and the remainder is \( 464 \).

\[
\boxed{7x^2 + 28x + 118 \text{ R } 464}
\]

---

Problem 7:


\[
(3x^4 - 5x^2 + 6) \div (x - 2)
\]

#### Synthetic Division Setup:
- Divisor: \( x - 2 \) → Root: \( r = 2 \)
- Coefficients of the dividend: \( 3, 0, -5, 0, 6 \) (Note: Include placeholders for \( x^3 \), \( x^1 \), and \( x^0 \)).

\[
\begin{array}{r|rrrrr}
2 & 3 & 0 & -5 & 0 & 6 \\
& & 6 & 12 & 14 & 28 \\
\hline
& 3 & 6 & 7 & 14 & 34 \\
\end{array}
\]

#### Explanation:
1. Bring down the leading coefficient: \( 3 \).
2. Multiply \( 3 \times 2 = 6 \) and add to the next coefficient: \( 0 + 6 = 6 \).
3. Multiply \( 6 \times 2 = 12 \) and add to the next coefficient: \( -5 + 12 = 7 \).
4. Multiply \( 7 \times 2 = 14 \) and add to the next coefficient: \( 0 + 14 = 14 \).
5. Multiply \( 14 \times 2 = 28 \) and add to the next coefficient: \( 6 + 28 = 34 \).

#### Result:
The quotient is \( 3x^3 + 6x^2 + 7x + 14 \) and the remainder is \( 34 \).

\[
\boxed{3x^3 + 6x^2 + 7x + 14 \text{ R } 34}
\]

---

Problem 8:


\[
(8x^2 - 5x + 6) \div (x - 2)
\]

#### Synthetic Division Setup:
- Divisor: \( x - 2 \) → Root: \( r = 2 \)
- Coefficients of the dividend: \( 8, -5, 6 \)

\[
\begin{array}{r|rrr}
2 & 8 & -5 & 6 \\
& & 16 & 22 \\
\hline
& 8 & 11 & 28 \\
\end{array}
\]

#### Explanation:
1. Bring down the leading coefficient: \( 8 \).
2. Multiply \( 8 \times 2 = 16 \) and add to the next coefficient: \( -5 + 16 = 11 \).
3. Multiply \( 11 \times 2 = 22 \) and add to the next coefficient: \( 6 + 22 = 28 \).

#### Result:
The quotient is \( 8x + 11 \) and the remainder is \( 28 \).

\[
\boxed{8x + 11 \text{ R } 28}
\]

---

Final Answers:


1. \( \boxed{x^2 + x - 2} \)
2. \( \boxed{x^3 + 7x + 1 \text{ R } -6} \)
3. \( \boxed{7x^2 - 14x + 32 \text{ R } -56} \)
4. \( \boxed{x^2 + 3x + 1} \)
5. \( \boxed{3x - 8 \text{ R } 20} \)
6. \( \boxed{7x^2 + 28x + 118 \text{ R } 464} \)
7. \( \boxed{3x^3 + 6x^2 + 7x + 14 \text{ R } 34} \)
8. \( \boxed{8x + 11 \text{ R } 28} \)
Parent Tip: Review the logic above to help your child master the concept of synthetic division worksheet answers.
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