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 of the fraction, and \( r(x) \) is the remainder.

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

Solution

To rewrite the fraction using polynomial long division, follow these steps: 1. Divide the first term of the numerator by the first term of the denominator. Divide \(4x^3\) by \(2x\) to get \(2x^2\). 2. Multiply the entire divisor by this result and subtract from the original polynomial. \[ \begin{align*} (2x^2)(2x + 1) &= 4x^3 + 2x^2 \end{align*} \] Now subtract: \[ \begin{align*} (4x^3 - 4x^2 + 9x + 6) - (4x^3 + 2x^2) &= -6x^2 + 9x + 6 \end{align*} \] 3. Repeat the process for the new polynomial term. Divide \(-6x^2\) by \(2x\) to get \(-3x\). 4. Multiply the divisor and subtract again. \[ \begin{align*} (-3x)(2x + 1) &= -6x^2 - 3x \end{align*} \] Subtract: \[ \begin{align*} (-6x^2 + 9x + 6) - (-6x^2 - 3x) &= 12x + 6 \end{align*} \] 5. Continue this process. Divide \(12x\) by \(2x\) to get \(6\). 6. Multiply and subtract once more. \[ \begin{align*} (6)(2x + 1) &= 12x + 6 \end{align*} \] Subtract: \[ \begin{align*} (12x + 6) - (12x + 6) &= 0 \end{align*} \] The complete division gives: \[ \begin{cases} q(x) = 2x^2 - 3x + 6 & \\ r(x) = 0 & \end{cases} \] So, the expression \(\frac{4x^3 - 4x^2 + 9x + 6}{2x + 1}\) is equal to \(2x^2 - 3x + 6\), with no remainder.