Use the Remainder Theorem to find which of the following is a factor of `2x^3 + 3x^2- 5x - 6.`
(i) x+1 (ii) 2x-1
(iii) x+2

Answer

To determine which of the given expressions is a factor of the polynomial \( f(x) = 2x^3 + 3x^2 - 5x - 6 \) using the Remainder Theorem, we will evaluate \( f(x) \) at specific values derived from the factors.

### Step-by-Step Solution:

1. **Identify the Polynomial:**
Let \( f(x) = 2x^3 + 3x^2 - 5x - 6 \).

2. **Check the first factor \( x + 1 \):**
- Set \( x + 1 = 0 \) to find \( x = -1 \).
- Calculate \( f(-1) \):
\[ f(-1) = 2(-1)^3 + 3(-1)^2 - 5(-1) - 6 \]
\[ = 2(-1) + 3(1) + 5 - 6 \]
\[ = -2 + 3 + 5 - 6 = 0 \]
- Since \( f(-1) = 0 \), \( x + 1 \) is a factor of \( f(x) \).

3. **Check the second factor \( 2x - 1 \):**
- Set \( 2x - 1 = 0 \) to find \( x = \frac{1}{2} \).
- Calculate \( f\left(\frac{1}{2}\right) \):
\[ f\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^3 + 3\left(\frac{1}{2}\right)^2 - 5\left(\frac{1}{2}\right) - 6 \]
\[ = 2\left(\frac{1}{8}\right) + 3\left(\frac{1}{4}\right) - \frac{5}{2} - 6 \]
\[ = \frac{1}{4} + \frac{3}{4} - \frac{5}{2} - 6 \]
\[ = 1 - \frac{5}{2} - 6 = 1 - 2.5 - 6 = -7.5 \neq 0 \]
- Since \( f\left(\frac{1}{2}\right) \neq 0 \), \( 2x - 1 \) is not a factor of \( f(x) \).

4. **Check the third factor \( x + 2 \):**
- Set \( x + 2 = 0 \) to find \( x = -2 \).
- Calculate \( f(-2) \):
\[ f(-2) = 2(-2)^3 + 3(-2)^2 - 5(-2) - 6 \]
\[ = 2(-8) + 3(4) + 10 - 6 \]
\[ = -16 + 12 + 10 - 6 \]
\[ = -16 + 12 + 10 - 6 = 0 \]
- Since \( f(-2) = 0 \), \( x + 2 \) is also a factor of \( f(x) \).

### Conclusion:
The factors of the polynomial \( 2x^3 + 3x^2 - 5x - 6 \) are \( x + 1 \) and \( x + 2 \). The factor \( 2x - 1 \) is not a factor.