Question

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

Answer 1

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

1. **Set up the long division**:
- Dividend: \( 3x^4 - 13x^3 + 7x^2 - x - 9 \)
- Divisor: \( x^2 - 4x + 1 \)

2. **Divide the leading term**:
- Divide \( 3x^4 \) by \( x^2 \) to get \( 3x^2 \).

3. **Multiply and subtract**:
- Multiply \( 3x^2 \) by the entire divisor:
\[
3x^2(x^2 - 4x + 1) = 3x^4 - 12x^3 + 3x^2
\]
- Subtract this from the original polynomial:
\[
(3x^4 - 13x^3 + 7x^2 - x - 9) - (3x^4 - 12x^3 + 3x^2) = -13x^3 + 12x^3 + 7x^2 - 3x^2 - x - 9
\]
Simplifying gives:
\[
-x^3 + 4x^2 - x - 9
\]

4. **Repeat the process**:
- Again, divide the leading term \( -x^3 \) by \( x^2 \):
\[
-x^3 \div x^2 = -x
\]
- Multiply \( -x \) by the divisor and subtract:
\[
-x(x^2 - 4x + 1) = -x^3 + 4x^2 - x
\]
- Subtract:
\[
(-x^3 + 4x^2 - x - 9) - (-x^3 + 4x^2 - x) = -9
\]

5. **Final step**:
- Now we have:
\[
-9
\]
- Since the degree of \(-9\) is less than the degree of the divisor, we stop here.

Putting it all together, we find that:

\[
\text{Quotient} = 3x^2 - x
\]
\[
\text{Remainder} = -9
\]

Thus,
\[
(3x^4 - 13x^3 + 7x^2 - x - 9) \div (x^2 - 4x + 1) = 3x^2 - x - \frac{9}{x^2 - 4x + 1}
\]

The quotient of the division is:

\[
\boxed{3x^2 - x}
\]