Algebra II A worksheet focusing on synthetic division and factoring polynomials.

Problem Analysis:

The task involves solving polynomial division problems using synthetic division. The goal is to determine whether a given binomial (e.g., \( x + 3 \) or \( x - 5 \)) is a factor of the polynomial by performing synthetic division and checking if the remainder is zero.

Step-by-Step Solution:

#### Problem 1:
Polynomial: \( P(x) = x^3 + 5x^2 + 3x + 9 \)
Divisor: \( x + 3 \)

1. Set up synthetic division:
- The divisor is \( x + 3 \), so the root is \( r = -3 \).
- Write down the coefficients of the polynomial \( P(x) \): \( 1, 5, 3, 9 \).

Synthetic division setup:
$$
\begin{array}{r|rrrr}
-3 & 1 & 5 & 3 & 9 \\
& & & & \\
\hline
& & & & \\
\end{array}
$$

2. Perform synthetic division:
- Bring down the leading coefficient \( 1 \).
- Multiply \( 1 \times (-3) = -3 \) and add to the next coefficient \( 5 \):
$$
5 + (-3) = 2
$$
- Multiply \( 2 \times (-3) = -6 \) and add to the next coefficient \( 3 \):
$$
3 + (-6) = -3
$$
- Multiply \( -3 \times (-3) = 9 \) and add to the next coefficient \( 9 \):
$$
9 + 9 = 18
$$

Updated synthetic division:
$$
\begin{array}{r|rrrr}
-3 & 1 & 5 & 3 & 9 \\
& & -3 & -6 & -9 \\
\hline
& 1 & 2 & -3 & 18 \\
\end{array}
$$

3. Interpret the result:
- The quotient is \( x^2 + 2x - 3 \).
- The remainder is \( 18 \).

4. Conclusion:
- Since the remainder is not zero (\( R = 18 \)), \( x + 3 \) is not a factor of \( P(x) \).

Answer for Problem 1:
$$
\boxed{\text{No, because the remainder is not 0.}}
$$

---

#### Problem 2:
Polynomial: \( P(x) = -x^3 + 9x^2 + 10x + 100 \)
Divisor: \( x - 5 \)

1. Set up synthetic division:
- The divisor is \( x - 5 \), so the root is \( r = 5 \).
- Write down the coefficients of the polynomial \( P(x) \): \( -1, 9, 10, 100 \).

Synthetic division setup:
$$
\begin{array}{r|rrrr}
5 & -1 & 9 & 10 & 100 \\
& & & & \\
\hline
& & & & \\
\end{array}
$$

2. Perform synthetic division:
- Bring down the leading coefficient \( -1 \).
- Multiply \( -1 \times 5 = -5 \) and add to the next coefficient \( 9 \):
$$
9 + (-5) = 4
$$
- Multiply \( 4 \times 5 = 20 \) and add to the next coefficient \( 10 \):
$$
10 + 20 = 30
$$
- Multiply \( 30 \times 5 = 150 \) and add to the next coefficient \( 100 \):
$$
100 + 150 = 250
$$

Updated synthetic division:
$$
\begin{array}{r|rrrr}
5 & -1 & 9 & 10 & 100 \\
& & -5 & 20 & 150 \\
\hline
& -1 & 4 & 30 & 250 \\
\end{array}
$$

3. Interpret the result:
- The quotient is \( -x^2 + 4x + 30 \).
- The remainder is \( 250 \).

4. Conclusion:
- Since the remainder is not zero (\( R = 250 \)), \( x - 5 \) is not a factor of \( P(x) \).

Answer for Problem 2:
$$
\boxed{\text{No, because the remainder is not 0.}}
$$

---

#### Problem 3:
Polynomial: \( P(x) = 2x^3 + x^2 - 3x - 6 \)
Divisor: \( x + 3 \)

1. Set up synthetic division:
- The divisor is \( x + 3 \), so the root is \( r = -3 \).
- Write down the coefficients of the polynomial \( P(x) \): \( 2, 1, -3, -6 \).

Synthetic division setup:
$$
\begin{array}{r|rrrr}
-3 & 2 & 1 & -3 & -6 \\
& & & & \\
\hline
& & & & \\
\end{array}
$$

2. Perform synthetic division:
- Bring down the leading coefficient \( 2 \).
- Multiply \( 2 \times (-3) = -6 \) and add to the next coefficient \( 1 \):
$$
1 + (-6) = -5
$$
- Multiply \( -5 \times (-3) = 15 \) and add to the next coefficient \( -3 \):
$$
-3 + 15 = 12
$$
- Multiply \( 12 \times (-3) = -36 \) and add to the next coefficient \( -6 \):
$$
-6 + (-36) = -42
$$

Updated synthetic division:
$$
\begin{array}{r|rrrr}
-3 & 2 & 1 & -3 & -6 \\
& & -6 & -5 & -36 \\
\hline
& 2 & -5 & 12 & -42 \\
\end{array}
$$

3. Interpret the result:
- The quotient is \( 2x^2 - 5x + 12 \).
- The remainder is \( -42 \).

4. Conclusion:
- Since the remainder is not zero (\( R = -42 \)), \( x + 3 \) is not a factor of \( P(x) \).

Answer for Problem 3:
$$
\boxed{\text{No, because the remainder is not 0.}}
$$

---

#### Problem 4:
Polynomial: \( P(x) = 2x^3 + x^2 - 3x - 6 \)
Divisor: \( x - 3 \)

1. Set up synthetic division:
- The divisor is \( x - 3 \), so the root is \( r = 3 \).
- Write down the coefficients of the polynomial \( P(x) \): \( 2, 1, -3, -6 \).

Synthetic division setup:
$$
\begin{array}{r|rrrr}
3 & 2 & 1 & -3 & -6 \\
& & & & \\
\hline
& & & & \\
\end{array}
$$

2. Perform synthetic division:
- Bring down the leading coefficient \( 2 \).
- Multiply \( 2 \times 3 = 6 \) and add to the next coefficient \( 1 \):
$$
1 + 6 = 7
$$
- Multiply \( 7 \times 3 = 21 \) and add to the next coefficient \( -3 \):
$$
-3 + 21 = 18
$$
- Multiply \( 18 \times 3 = 54 \) and add to the next coefficient \( -6 \):
$$
-6 + 54 = 48
$$

Updated synthetic division:
$$
\begin{array}{r|rrrr}
3 & 2 & 1 & -3 & -6 \\
& & 6 & 21 & 54 \\
\hline
& 2 & 7 & 18 & 48 \\
\end{array}
$$

3. Interpret the result:
- The quotient is \( 2x^2 + 7x + 18 \).
- The remainder is \( 48 \).

4. Conclusion:
- Since the remainder is not zero (\( R = 48 \)), \( x - 3 \) is not a factor of \( P(x) \).

Answer for Problem 4:
$$
\boxed{\text{No, because the remainder is not 0.}}
$$

---

Final Answers:

1. \( \boxed{\text{No, because the remainder is not 0.}} \)
2. \( \boxed{\text{No, because the remainder is not 0.}} \)
3. \( \boxed{\text{No, because the remainder is not 0.}} \)
4. \( \boxed{\text{No, because the remainder is not 0.}} \)