What must be subtracted from `x^3 - 6x^2 +13x – 6` so that the resulting polynomial is exactly divisible `x^(2) +x+1`?

Answer

To solve the problem of what must be subtracted from the polynomial \(x^3 - 6x^2 + 13x - 6\) so that the resulting polynomial is exactly divisible by \(x^2 + x + 1\), we can follow these steps:

### Step-by-Step Solution:

1. **Define the Polynomials**:
Let \(f(x) = x^3 - 6x^2 + 13x - 6\) and \(g(x) = x^2 + x + 1\).

2. **Perform Polynomial Long Division**:
We need to divide \(f(x)\) by \(g(x)\) to find the remainder.

- Divide the leading term of \(f(x)\) by the leading term of \(g(x)\):
\[ \frac{x^3}{x^2} = x \]
- Multiply \(g(x)\) by \(x\):
\[ x \cdot (x^2 + x + 1) = x^3 + x^2 + x \]
- Subtract this from \(f(x)\):
\[ (x^3 - 6x^2 + 13x - 6) - (x^3 + x^2 + x) = -7x^2 + 12x - 6 \]

3. **Continue the Division**:
- Now, divide the leading term of the new polynomial \(-7x^2\) by the leading term of \(g(x)\):
\[ \frac{-7x^2}{x^2} = -7 \]
- Multiply \(g(x)\) by \(-7\):
\[ -7 \cdot (x^2 + x + 1) = -7x^2 - 7x - 7 \]
- Subtract this from \(-7x^2 + 12x - 6\):
\[ (-7x^2 + 12x - 6) - (-7x^2 - 7x - 7) = 19x + 1 \]

4. **Find the Remainder**:
The remainder from the division is \(19x + 1\).

5. **Determine What to Subtract**:
For \(f(x)\) to be exactly divisible by \(g(x)\), the remainder must be zero. Therefore, we need to subtract the remainder from \(f(x)\):
\[ 19x + 1 \]

### Conclusion:
Thus, the polynomial that must be subtracted from \(x^3 - 6x^2 + 13x - 6\) is \(19x + 1\).