If `8x^(4)-8x^(2)+7` is divided by `2x+1`, the remainder is

Options

A. `(11)/(2)`
B. `(13)/(2)`
C. `(15)/(2)`
D. `(17)/(2)`

Answer

To find the remainder when the polynomial \( 8x^4 - 8x^2 + 7 \) is divided by \( 2x + 1 \), we can use the Remainder Theorem. According to the theorem, the remainder of the division of a polynomial \( f(x) \) by a linear divisor \( ax + b \) is given by \( f(-\frac{b}{a}) \).

### Step-by-Step Solution:

1. **Identify the polynomial and the divisor**:
- The polynomial is \( f(x) = 8x^4 - 8x^2 + 7 \).
- The divisor is \( 2x + 1 \).

2. **Find the value of \( x \)**:
- Set \( 2x + 1 = 0 \) to find the value of \( x \) at which to evaluate the polynomial.
- Solving for \( x \):
\[ 2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2} \]

3. **Evaluate the polynomial at \( x = -\frac{1}{2} \)**:
- Substitute \( x = -\frac{1}{2} \) into the polynomial \( f(x) \):
\[ f\left(-\frac{1}{2}\right) = 8\left(-\frac{1}{2}\right)^4 - 8\left(-\frac{1}{2}\right)^2 + 7 \]

4. **Calculate \( \left(-\frac{1}{2}\right)^4 \) and \( \left(-\frac{1}{2}\right)^2 \)**:
- \( \left(-\frac{1}{2}\right)^4 = \frac{1}{16} \)
- \( \left(-\frac{1}{2}\right)^2 = \frac{1}{4} \)

5. **Substitute these values back into the polynomial**:
\[ f\left(-\frac{1}{2}\right) = 8 \cdot \frac{1}{16} - 8 \cdot \frac{1}{4} + 7 \]

6. **Simplify the expression**:
- Calculate \( 8 \cdot \frac{1}{16} = \frac{8}{16} = \frac{1}{2} \)
- Calculate \( 8 \cdot \frac{1}{4} = 2 \)
- Now substitute these values:
\[ f\left(-\frac{1}{2}\right) = \frac{1}{2} - 2 + 7 \]

7. **Combine the terms**:
\[ f\left(-\frac{1}{2}\right) = \frac{1}{2} - \frac{4}{2} + \frac{14}{2} = \frac{1 - 4 + 14}{2} = \frac{11}{2} \]

8. **Conclusion**:
- The remainder when \( 8x^4 - 8x^2 + 7 \) is divided by \( 2x + 1 \) is \( \frac{11}{2} \).

### Final Answer:
The remainder is \( \frac{11}{2} \).