If `x ^(3) - 7x + 4` is divided by `(x-1), (x +2) and (2x +1),` then the find the respective remainders.

Options

A. `-2, 10, 7 (3)/(8)`
B. `-4, -5, (8)/(9)`
C. `- 4, -6, - 8`
D. `-2, - 9, 7`

Answer

To find the remainders when the polynomial \( P(x) = x^3 - 7x + 4 \) is divided by \( (x - 1) \), \( (x + 2) \), and \( (2x + 1) \), we can use the Remainder Theorem. According to the theorem, the remainder of the division of a polynomial \( P(x) \) by \( (x - c) \) is \( P(c) \).

### Step 1: Find the remainder when \( P(x) \) is divided by \( (x - 1) \)

1. Substitute \( x = 1 \) into \( P(x) \):
\[ P(1) = 1^3 - 7(1) + 4 = 1 - 7 + 4 = -2 \]
Therefore, the remainder when \( P(x) \) is divided by \( (x - 1) \) is \( -2 \).

### Step 2: Find the remainder when \( P(x) \) is divided by \( (x + 2) \)

1. Substitute \( x = -2 \) into \( P(x) \):
\[ P(-2) = (-2)^3 - 7(-2) + 4 = -8 + 14 + 4 = 10 \]
Therefore, the remainder when \( P(x) \) is divided by \( (x + 2) \) is \( 10 \).

### Step 3: Find the remainder when \( P(x) \) is divided by \( (2x + 1) \)

1. Substitute \( x = -\frac{1}{2} \) into \( P(x) \):
\[ P\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^3 - 7\left(-\frac{1}{2}\right) + 4 = -\frac{1}{8} + \frac{7}{2} + 4 \]
Convert \( \frac{7}{2} \) and \( 4 \) to eighths:
\[ \frac{7}{2} = \frac{28}{8}, \quad 4 = \frac{32}{8} \]
Now combine:
\[ P\left(-\frac{1}{2}\right) = -\frac{1}{8} + \frac{28}{8} + \frac{32}{8} = \frac{-1 + 28 + 32}{8} = \frac{59}{8} \]
Therefore, the remainder when \( P(x) \) is divided by \( (2x + 1) \) is \( \frac{59}{8} \).

### Final Remainders

1. Remainder when divided by \( (x - 1) \): \( -2 \)
2. Remainder when divided by \( (x + 2) \): \( 10 \)
3. Remainder when divided by \( (2x + 1) \): \( \frac{59}{8} \)

### Summary of Remainders
- Remainder for \( (x - 1) \): \( -2 \)
- Remainder for \( (x + 2) \): \( 10 \)
- Remainder for \( (2x + 1) \): \( \frac{59}{8} \)

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