Synthetic Division Worksheets - Math Monks - Free Printable

To solve the given problems using synthetic division, we need to follow these steps:

1. Rewrite the divisor in the form \( x - c \): If the divisor is not already in this form, adjust it accordingly.
2. Set up the synthetic division: Write down the coefficients of the dividend and use the root \( c \) from the divisor.
3. Perform the synthetic division: Bring down the leading coefficient, multiply, add, and repeat until all coefficients are processed.
4. Interpret the result: The numbers on the bottom row represent the coefficients of the quotient polynomial, and the last number is the remainder.

Let's solve each problem step by step.

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Problem 1: \( (6x^3 + 23x^2 + 10x + 1) \div (3x + 1) \)

#### Step 1: Rewrite the divisor
The divisor is \( 3x + 1 \). To use synthetic division, we need it in the form \( x - c \). Divide the entire divisor by 3:
\[
3x + 1 = 3(x + \frac{1}{3})
\]
So, we can rewrite the problem as:
\[
\frac{6x^3 + 23x^2 + 10x + 1}{3(x + \frac{1}{3})} = \frac{1}{3} \cdot \frac{6x^3 + 23x^2 + 10x + 1}{x + \frac{1}{3}}
\]
Now, we will perform synthetic division with \( c = -\frac{1}{3} \).

#### Step 2: Set up the synthetic division
The coefficients of the dividend \( 6x^3 + 23x^2 + 10x + 1 \) are \( 6, 23, 10, 1 \). The root \( c = -\frac{1}{3} \).

\[
\begin{array}{r|rrrr}
-\frac{1}{3} & 6 & 23 & 10 & 1 \\
& & -2 & -7 & -1 \\
\hline
& 6 & 21 & 3 & 0 \\
\end{array}
\]

#### Step 3: Interpret the result
The numbers on the bottom row are \( 6, 21, 3, 0 \). The quotient is \( 6x^2 + 21x + 3 \), and the remainder is 0. Since we initially factored out \( \frac{1}{3} \), the final quotient is:
\[
\frac{1}{3}(6x^2 + 21x + 3) = 2x^2 + 7x + 1
\]

#### Final Answer:
\[
\boxed{2x^2 + 7x + 1}
\]

---

Problem 2: \( (x^3 + 4x^2 - 5x + 3) \div (x + 2) \)

#### Step 1: Rewrite the divisor
The divisor is \( x + 2 \), so \( c = -2 \).

#### Step 2: Set up the synthetic division
The coefficients of the dividend \( x^3 + 4x^2 - 5x + 3 \) are \( 1, 4, -5, 3 \).

\[
\begin{array}{r|rrrr}
-2 & 1 & 4 & -5 & 3 \\
& & -2 & -4 & 18 \\
\hline
& 1 & 2 & -9 & 21 \\
\end{array}
\]

#### Step 3: Interpret the result
The numbers on the bottom row are \( 1, 2, -9, 21 \). The quotient is \( x^2 + 2x - 9 \), and the remainder is 21.

#### Final Answer:
\[
\boxed{x^2 + 2x - 9 + \frac{21}{x + 2}}
\]

---

Problem 3: \( (14x^2 + 69x + 27) \div (7x + 3) \)

#### Step 1: Rewrite the divisor
The divisor is \( 7x + 3 \). To use synthetic division, we need it in the form \( x - c \). Divide the entire divisor by 7:
\[
7x + 3 = 7(x + \frac{3}{7})
\]
So, we can rewrite the problem as:
\[
\frac{14x^2 + 69x + 27}{7(x + \frac{3}{7})} = \frac{1}{7} \cdot \frac{14x^2 + 69x + 27}{x + \frac{3}{7}}
\]
Now, we will perform synthetic division with \( c = -\frac{3}{7} \).

#### Step 2: Set up the synthetic division
The coefficients of the dividend \( 14x^2 + 69x + 27 \) are \( 14, 69, 27 \). The root \( c = -\frac{3}{7} \).

\[
\begin{array}{r|rrr}
-\frac{3}{7} & 14 & 69 & 27 \\
& & -6 & -63 \\
\hline
& 14 & 63 & 0 \\
\end{array}
\]

#### Step 3: Interpret the result
The numbers on the bottom row are \( 14, 63, 0 \). The quotient is \( 14x + 63 \), and the remainder is 0. Since we initially factored out \( \frac{1}{7} \), the final quotient is:
\[
\frac{1}{7}(14x + 63) = 2x + 9
\]

#### Final Answer:
\[
\boxed{2x + 9}
\]

---

Problem 4: \( (12x^2 - 39x + 30) \div (4x - 5) \)

#### Step 1: Rewrite the divisor
The divisor is \( 4x - 5 \). To use synthetic division, we need it in the form \( x - c \). Divide the entire divisor by 4:
\[
4x - 5 = 4(x - \frac{5}{4})
\]
So, we can rewrite the problem as:
\[
\frac{12x^2 - 39x + 30}{4(x - \frac{5}{4})} = \frac{1}{4} \cdot \frac{12x^2 - 39x + 30}{x - \frac{5}{4}}
\]
Now, we will perform synthetic division with \( c = \frac{5}{4} \).

#### Step 2: Set up the synthetic division
The coefficients of the dividend \( 12x^2 - 39x + 30 \) are \( 12, -39, 30 \). The root \( c = \frac{5}{4} \).

\[
\begin{array}{r|rrr}
\frac{5}{4} & 12 & -39 & 30 \\
& & 15 & -30 \\
\hline
& 12 & -24 & 0 \\
\end{array}
\]

#### Step 3: Interpret the result
The numbers on the bottom row are \( 12, -24, 0 \). The quotient is \( 12x - 24 \), and the remainder is 0. Since we initially factored out \( \frac{1}{4} \), the final quotient is:
\[
\frac{1}{4}(12x - 24) = 3x - 6
\]

#### Final Answer:
\[
\boxed{3x - 6}
\]

---

Problem 5: \( (7x^2 + 25x + 12) \div (x - 3) \)

#### Step 1: Rewrite the divisor
The divisor is \( x - 3 \), so \( c = 3 \).

#### Step 2: Set up the synthetic division
The coefficients of the dividend \( 7x^2 + 25x + 12 \) are \( 7, 25, 12 \).

\[
\begin{array}{r|rrr}
3 & 7 & 25 & 12 \\
& & 21 & 144 \\
\hline
& 7 & 46 & 156 \\
\end{array}
\]

#### Step 3: Interpret the result
The numbers on the bottom row are \( 7, 46, 156 \). The quotient is \( 7x + 46 \), and the remainder is 156.

#### Final Answer:
\[
\boxed{7x + 46 + \frac{156}{x - 3}}
\]

---

Problem 6: \( (5x^3 + 4x^2 - 45x - 36) \div (x - 9) \)

#### Step 1: Rewrite the divisor
The divisor is \( x - 9 \), so \( c = 9 \).

#### Step 2: Set up the synthetic division
The coefficients of the dividend \( 5x^3 + 4x^2 - 45x - 36 \) are \( 5, 4, -45, -36 \).

\[
\begin{array}{r|rrrr}
9 & 5 & 4 & -45 & -36 \\
& & 45 & 441 & 3576 \\
\hline
& 5 & 49 & 396 & 3540 \\
\end{array}
\]

#### Step 3: Interpret the result
The numbers on the bottom row are \( 5, 49, 396, 3540 \). The quotient is \( 5x^2 + 49x + 396 \), and the remainder is 3540.

#### Final Answer:
\[
\boxed{5x^2 + 49x + 396 + \frac{3540}{x - 9}}
\]

---

Problem 7: \( (4x^2 + 3x - 8) \div (x + 2) \)

#### Step 1: Rewrite the divisor
The divisor is \( x + 2 \), so \( c = -2 \).

#### Step 2: Set up the synthetic division
The coefficients of the dividend \( 4x^2 + 3x - 8 \) are \( 4, 3, -8 \).

\[
\begin{array}{r|rrr}
-2 & 4 & 3 & -8 \\
& & -8 & 10 \\
\hline
& 4 & -5 & 2 \\
\end{array}
\]

#### Step 3: Interpret the result
The numbers on the bottom row are \( 4, -5, 2 \). The quotient is \( 4x - 5 \), and the remainder is 2.

#### Final Answer:
\[
\boxed{4x - 5 + \frac{2}{x + 2}}
\]

---

Problem 8: \( (x^2 - x - 21) \div (x + 4) \)

#### Step 1: Rewrite the divisor
The divisor is \( x + 4 \), so \( c = -4 \).

#### Step 2: Set up the synthetic division
The coefficients of the dividend \( x^2 - x - 21 \) are \( 1, -1, -21 \).

\[
\begin{array}{r|rrr}
-4 & 1 & -1 & -21 \\
& & -4 & 20 \\
\hline
& 1 & -5 & -1 \\
\end{array}
\]

#### Step 3: Interpret the result
The numbers on the bottom row are \( 1, -5, -1 \). The quotient is \( x - 5 \), and the remainder is -1.

#### Final Answer:
\[
\boxed{x - 5 - \frac{1}{x + 4}}
\]

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Final Answers for All Problems:

1. \( \boxed{2x^2 + 7x + 1} \)
2. \( \boxed{x^2 + 2x - 9 + \frac{21}{x + 2}} \)
3. \( \boxed{2x + 9} \)
4. \( \boxed{3x - 6} \)
5. \( \boxed{7x + 46 + \frac{156}{x - 3}} \)
6. \( \boxed{5x^2 + 49x + 396 + \frac{3540}{x - 9}} \)
7. \( \boxed{4x - 5 + \frac{2}{x + 2}} \)
8. \( \boxed{x - 5 - \frac{1}{x + 4}} \)