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:

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