Question

What is the quotient of (3x4−13x3+7x2−x−9)÷(x2−4x+1)

Answer 1

To find the quotient of the polynomial \(3x^4 - 13x^3 + 7x^2 - x - 9\) divided by \(x^2 - 4x + 1\), we can use polynomial long division.

1. **Setup the long division:**
- Divide \(3x^4\) by \(x^2\) to get \(3x^2\).
- Multiply \(3x^2\) by \(x^2 - 4x + 1\):
\[
3x^2(x^2 - 4x + 1) = 3x^4 - 12x^3 + 3x^2
\]
- Subtract this from the original polynomial:
\[
(3x^4 - 13x^3 + 7x^2 - x - 9) - (3x^4 - 12x^3 + 3x^2) = -x^3 + 4x^2 - x - 9
\]

2. **Repeat the process:**
- Divide \(-x^3\) by \(x^2\) to get \(-x\).
- Multiply \(-x\) by \(x^2 - 4x + 1\):
\[
-x(x^2 - 4x + 1) = -x^3 + 4x^2 - x
\]
- Subtract:
\[
(-x^3 + 4x^2 - x - 9) - (-x^3 + 4x^2 - x) = -9
\]

3. **Final step:**
- There are no more terms to bring down, and we have reached the end of the division process. The division leaves us with a remainder of \(-9\).

Putting it all together, we have:
\[
\text{Quotient} = 3x^2 - x \quad \text{and} \quad \text{Remainder} = -9
\]

Thus, the result of the division is:
\[
3x^2 - x + \frac{-9}{x^2 - 4x + 1}
\]

**Final Answer:**
The quotient of \((3x^4 - 13x^3 + 7x^2 - x - 9)÷(x^2 - 4x + 1)\) is \(3x^2 - x\) with a remainder of \(-9\) (or \(3x^2 - x - \frac{9}{x^2 - 4x + 1}\)).