Question: Use polynomial long division to rewrite the following fraction in the form \( q(x) +…

Use polynomial long division to rewrite the following fraction in the form \( q(x) + \frac{r(x)}{d(x)} \), where \( d(x) \) is the denominator, \( q(x) \) is the quotient, and \( r(x) \) is the remainder.

\[ \frac{4x^3 - 4x^2 + 9x + 6}{2x + 1} \]

Answer: 6 Points

\[ 2x^2 - 3x + 6 \]

Solution

To perform polynomial long division, we divide \(4x^3 - 4x^2 + 9x + 6\) by \(2x + 1\). First, divide the leading term of the dividend \(4x^3\) by the leading term of the divisor \(2x\). \[ \frac{4x^3}{2x} = 2x^2 \] Multiply \(2x^2\) by the divisor \(2x + 1\) and subtract from the original dividend. \[ (2x^2)(2x + 1) = 4x^3 + 2x^2 \] Subtract this from \(4x^3 - 4x^2 + 9x + 6\). \[ \begin{align*} (4x^3 - 4x^2 + 9x + 6) - (4x^3 + 2x^2) &= -6x^2 + 9x + 6 \end{align*} \] Next, divide \(-6x^2\) by \(2x\). \[ \frac{-6x^2}{2x} = -3x \] Multiply \(-3x\) by the divisor \(2x + 1\). \[ (-3x)(2x + 1) = -6x^2 - 3x \] Subtract from \(-6x^2 + 9x + 6\). \[ \begin{align*} (-6x^2 + 9x + 6) - (-6x^2 - 3x) &= 12x + 6 \end{align*} \] Now, divide \(12x\) by \(2x\). \[ \frac{12x}{2x} = 6 \] Multiply \(6\) by \(2x + 1\). \[ 6(2x + 1) = 12x + 6 \] Subtract from \(12x + 6\). \[ \begin{align*} (12x + 6) - (12x + 6) &= 0 \end{align*} \] The quotient is \(2x^2 - 3x + 6\) and the remainder is 0. So, \[ \frac{4x^3 - 4x^2 + 9x + 6}{2x + 1} = 2x^2 - 3x + 6 \]