What must be added to the polynomial `2x^(3)-3x^(2)-8x`, so that it leaves a remainder 10 when divided by `2x +1` ?

Answer

To solve the problem, we need to determine what must be added to the polynomial \( f(x) = 2x^3 - 3x^2 - 8x \) so that when it is divided by \( 2x + 1 \), the remainder is 10.

### Step-by-Step Solution:

1. **Define the Polynomial and the Addition**:
Let \( b \) be the value that we need to add to the polynomial. Thus, we can express the new polynomial as:
\[ f'(x) = 2x^3 - 3x^2 - 8x + b \]

2. **Use Polynomial Division**:
We will divide \( f'(x) \) by \( 2x + 1 \) and find the remainder. The remainder when dividing by a linear polynomial \( ax + b \) can be found using the Remainder Theorem, which states that the remainder of \( f(x) \) when divided by \( ax + b \) is \( f(-\frac{b}{a}) \).

Here, \( a = 2 \) and \( b = 1 \), so we need to evaluate \( f'(-\frac{1}{2}) \).

3. **Calculate \( f'(-\frac{1}{2}) \)**:
Substitute \( x = -\frac{1}{2} \) into \( f'(x) \):
\[ f'(-\frac{1}{2}) = 2\left(-\frac{1}{2}\right)^3 - 3\left(-\frac{1}{2}\right)^2 - 8\left(-\frac{1}{2}\right) + b \]
Calculate each term:
\[ = 2\left(-\frac{1}{8}\right) - 3\left(\frac{1}{4}\right) + 4 + b \]
\[ = -\frac{1}{4} - \frac{3}{4} + 4 + b \]
\[ = -1 + 4 + b \]
\[ = 3 + b \]

4. **Set the Remainder Equal to 10**:
According to the problem, we want the remainder to be 10:
\[ 3 + b = 10 \]

5. **Solve for \( b \)**:
To find \( b \), subtract 3 from both sides:
\[ b = 10 - 3 \]
\[ b = 7 \]

### Conclusion:
The value that must be added to the polynomial \( 2x^3 - 3x^2 - 8x \) so that it leaves a remainder of 10 when divided by \( 2x + 1 \) is \( \boxed{7} \).