Worksheet 2.1 & 2.2 on Remainder Theorem and Factor Theorem, with exercises for finding remainders and factors of a polynomial.

Problem Analysis:

The worksheet involves the Remainder Theorem and the Factor Theorem. Let's break down the problem step by step.

#### 1. Remainder Theorem
The Remainder Theorem states that when a polynomial \( f(x) \) is divided by \( x - a \), the remainder is \( f(a) \). We are given the polynomial:
\[ f(x) = 2x^3 + 3x^2 - 17x - 30 \]
and need to find the remainder when \( f(x) \) is divided by each of the following:
- (a) \( x - 1 \)
- (b) \( x - 2 \)
- (c) \( x - 3 \)
- (d) \( x + 1 \)
- (e) \( x + 2 \)
- (f) \( x + 3 \)

#### 2. Factor Theorem
The Factor Theorem states that if \( f(a) = 0 \), then \( x - a \) is a factor of \( f(x) \). Using the results from part 1, we will determine which of the given binomials are factors of \( f(x) \).

#### 3. Polynomial Division
For the binomials identified as factors in part 2, we will perform polynomial division of \( f(x) \) by \( x - a \) and fully factor the result.

---

Solution:

#### Part 1: Finding the Remainder

We use the Remainder Theorem to evaluate \( f(a) \) for each case.

Given:
\[ f(x) = 2x^3 + 3x^2 - 17x - 30 \]

##### (a) \( x - 1 \):
Set \( a = 1 \):
\[ f(1) = 2(1)^3 + 3(1)^2 - 17(1) - 30 \]
\[ f(1) = 2 + 3 - 17 - 30 \]
\[ f(1) = -42 \]

##### (b) \( x - 2 \):
Set \( a = 2 \):
\[ f(2) = 2(2)^3 + 3(2)^2 - 17(2) - 30 \]
\[ f(2) = 2(8) + 3(4) - 34 - 30 \]
\[ f(2) = 16 + 12 - 34 - 30 \]
\[ f(2) = -36 \]

##### (c) \( x - 3 \):
Set \( a = 3 \):
\[ f(3) = 2(3)^3 + 3(3)^2 - 17(3) - 30 \]
\[ f(3) = 2(27) + 3(9) - 51 - 30 \]
\[ f(3) = 54 + 27 - 51 - 30 \]
\[ f(3) = 0 \]

##### (d) \( x + 1 \):
Set \( a = -1 \):
\[ f(-1) = 2(-1)^3 + 3(-1)^2 - 17(-1) - 30 \]
\[ f(-1) = 2(-1) + 3(1) + 17 - 30 \]
\[ f(-1) = -2 + 3 + 17 - 30 \]
\[ f(-1) = -12 \]

##### (e) \( x + 2 \):
Set \( a = -2 \):
\[ f(-2) = 2(-2)^3 + 3(-2)^2 - 17(-2) - 30 \]
\[ f(-2) = 2(-8) + 3(4) + 34 - 30 \]
\[ f(-2) = -16 + 12 + 34 - 30 \]
\[ f(-2) = 0 \]

##### (f) \( x + 3 \):
Set \( a = -3 \):
\[ f(-3) = 2(-3)^3 + 3(-3)^2 - 17(-3) - 30 \]
\[ f(-3) = 2(-27) + 3(9) + 51 - 30 \]
\[ f(-3) = -54 + 27 + 51 - 30 \]
\[ f(-3) = 0 \]

#### Summary of Remainders:
\[
\begin{aligned}
&\text{(a)} \quad x - 1: \quad f(1) = -42 \\
&\text{(b)} \quad x - 2: \quad f(2) = -36 \\
&\text{(c)} \quad x - 3: \quad f(3) = 0 \\
&\text{(d)} \quad x + 1: \quad f(-1) = -12 \\
&\text{(e)} \quad x + 2: \quad f(-2) = 0 \\
&\text{(f)} \quad x + 3: \quad f(-3) = 0
\end{aligned}
\]

---

#### Part 2: Identifying Factors Using the Factor Theorem

From the results in Part 1, we see that:
- \( f(3) = 0 \), so \( x - 3 \) is a factor.
- \( f(-2) = 0 \), so \( x + 2 \) is a factor.
- \( f(-3) = 0 \), so \( x + 3 \) is a factor.

Thus, the factors are:
\[ x - 3, \quad x + 2, \quad x + 3 \]

---

#### Part 3: Polynomial Division

We now divide \( f(x) = 2x^3 + 3x^2 - 17x - 30 \) by each of the factors \( x - 3 \), \( x + 2 \), and \( x + 3 \).

##### (i) Dividing by \( x - 3 \):

Using synthetic division with \( a = 3 \):

\[
\begin{array}{r|rrrr}
3 & 2 & 3 & -17 & -30 \\
& & 6 & 27 & 30 \\
\hline
& 2 & 9 & 10 & 0 \\
\end{array}
\]

The quotient is \( 2x^2 + 9x + 10 \), and the remainder is 0. Thus:
\[ f(x) = (x - 3)(2x^2 + 9x + 10) \]

##### (ii) Dividing by \( x + 2 \):

Using synthetic division with \( a = -2 \):

\[
\begin{array}{r|rrrr}
-2 & 2 & 3 & -17 & -30 \\
& & -4 & 2 & -30 \\
\hline
& 2 & -1 & -15 & 0 \\
\end{array}
\]

The quotient is \( 2x^2 - x - 15 \), and the remainder is 0. Thus:
\[ f(x) = (x + 2)(2x^2 - x - 15) \]

##### (iii) Dividing by \( x + 3 \):

Using synthetic division with \( a = -3 \):

\[
\begin{array}{r|rrrr}
-3 & 2 & 3 & -17 & -30 \\
& & -6 & 9 & 24 \\
\hline
& 2 & -3 & -8 & 0 \\
\end{array}
\]

The quotient is \( 2x^2 - 3x - 8 \), and the remainder is 0. Thus:
\[ f(x) = (x + 3)(2x^2 - 3x - 8) \]

---

Final Answer:

\[
\boxed{x - 3, \, x + 2, \, x + 3}
\]