Find the quotient and the remainder (if any), when :
`3x^(4) + 6x^(3) - 6x^(2) + 2x - 7` is divided by x - 3.

Answer

To find the quotient and the remainder when the polynomial \(3x^4 + 6x^3 - 6x^2 + 2x - 7\) is divided by \(x - 3\), we will use polynomial long division.

### Step-by-Step Solution:

1. **Write the Polynomial in Descending Order**:
The polynomial is already in descending order:
\[ 3x^4 + 6x^3 - 6x^2 + 2x - 7 \]

2. **Divide the Leading Terms**:
Divide the leading term of the polynomial \(3x^4\) by the leading term of the divisor \(x\):
\[ \frac{3x^4}{x} = 3x^3 \]
This is the first term of our quotient.

3. **Multiply and Subtract**:
Multiply \(3x^3\) by the entire divisor \(x - 3\):
\[ 3x^3 \cdot (x - 3) = 3x^4 - 9x^3 \]
Now, subtract this from the original polynomial:
\[ (3x^4 + 6x^3) - (3x^4 - 9x^3) = 6x^3 + 9x^3 = 15x^3 \]
So, we have:
\[ 15x^3 - 6x^2 + 2x - 7 \]

4. **Repeat the Process**:
Now, divide the leading term \(15x^3\) by \(x\):
\[ \frac{15x^3}{x} = 15x^2 \]
Multiply \(15x^2\) by the divisor:
\[ 15x^2 \cdot (x - 3) = 15x^3 - 45x^2 \]
Subtract:
\[ (15x^3 - 6x^2) - (15x^3 - 45x^2) = -6x^2 + 45x^2 = 39x^2 \]
So now we have:
\[ 39x^2 + 2x - 7 \]

5. **Continue Dividing**:
Divide \(39x^2\) by \(x\):
\[ \frac{39x^2}{x} = 39x \]
Multiply \(39x\) by the divisor:
\[ 39x \cdot (x - 3) = 39x^2 - 117x \]
Subtract:
\[ (39x^2 + 2x) - (39x^2 - 117x) = 2x + 117x = 119x \]
So we have:
\[ 119x - 7 \]

6. **Final Division**:
Divide \(119x\) by \(x\):
\[ \frac{119x}{x} = 119 \]
Multiply \(119\) by the divisor:
\[ 119 \cdot (x - 3) = 119x - 357 \]
Subtract:
\[ (119x - 7) - (119x - 357) = -7 + 357 = 350 \]

7. **Conclusion**:
The quotient is:
\[ 3x^3 + 15x^2 + 39x + 119 \]
The remainder is:
\[ 350 \]

### Final Answer:
- **Quotient**: \(3x^3 + 15x^2 + 39x + 119\)
- **Remainder**: \(350\)