What should be subtracted from f(x) = `6x^(3) + 11x^(2) – 39x - 65` so that f(x) is exactly divisible by `x^(2) + x - 1`?

Options

A. 38 x + 60
B. `-38 -60 `
C. `-19x - 30`
D. `9x+10`

Answer

To find out what should be subtracted from the polynomial \( f(x) = 6x^3 + 11x^2 - 39x - 65 \) so that it is exactly divisible by \( g(x) = x^2 + x - 1 \), we need to perform polynomial long division and find the remainder.

### Step-by-step Solution:

1. **Set up the division**:
We want to divide \( f(x) \) by \( g(x) \):
\[ f(x) = 6x^3 + 11x^2 - 39x - 65 \]
\[ g(x) = x^2 + x - 1 \]

2. **Perform polynomial long division**:
- Divide the leading term of \( f(x) \) (which is \( 6x^3 \)) by the leading term of \( g(x) \) (which is \( x^2 \)):
\[ \frac{6x^3}{x^2} = 6x \]
- Multiply \( g(x) \) by \( 6x \):
\[ 6x \cdot (x^2 + x - 1) = 6x^3 + 6x^2 - 6x \]
- Subtract this from \( f(x) \):
\[ (6x^3 + 11x^2 - 39x - 65) - (6x^3 + 6x^2 - 6x) = (11x^2 - 6x^2) + (-39x + 6x) - 65 = 5x^2 - 33x - 65 \]

3. **Repeat the process**:
- Now, divide the leading term of the new polynomial \( 5x^2 \) by the leading term of \( g(x) \):
\[ \frac{5x^2}{x^2} = 5 \]
- Multiply \( g(x) \) by \( 5 \):
\[ 5 \cdot (x^2 + x - 1) = 5x^2 + 5x - 5 \]
- Subtract this from \( 5x^2 - 33x - 65 \):
\[ (5x^2 - 33x - 65) - (5x^2 + 5x - 5) = (-33x - 5x) + (-65 + 5) = -38x - 60 \]

4. **Find the remainder**:
The remainder after the division is:
\[ R(x) = -38x - 60 \]

5. **Determine what to subtract**:
To make \( f(x) \) exactly divisible by \( g(x) \), we need to subtract the remainder:
\[ \text{What to subtract} = -(-38x - 60) = 38x + 60 \]

### Final Answer:
Therefore, the expression that should be subtracted from \( f(x) \) is:
\[ \boxed{38x + 60} \]