Question

using long division, what is the quotient of this expression?\\(dfrac{3x^4 - 2x^3 - x - 4}{x^2 + 2}\\)\\(\text{a. } 3x^2 - 2x - 5 + dfrac{3x + 6}{x^2 + 2}\\)\\(\text{b. } 3x^2 + 2x + dfrac{3x - 4}{x^2 + 2}\\)\\(\text{c. } 3x^2 - 2x - 6 + dfrac{3x + 8}{x^2 + 2}\\)\\(\text{d. } 3x^2 + 2x - dfrac{5x - 8}{x^2 + 2}\\)

Explanation:

Step1: Divide the leading term

Divide the leading term of the numerator \(3x^4\) by the leading term of the denominator \(x^2\), we get \(3x^2\). Multiply the denominator \(x^2 + 2\) by \(3x^2\) to get \(3x^4+6x^2\). Subtract this from the numerator:
\[

$$\begin{align*} &(3x^4 - 2x^3 - x - 4)-(3x^4 + 6x^2)\\ =&3x^4 - 2x^3 - x - 4 - 3x^4 - 6x^2\\ =& - 2x^3 - 6x^2 - x - 4 \end{align*}$$

\]

Step2: Divide the new leading term

Divide the leading term of the new polynomial \(-2x^3\) by \(x^2\), we get \(-2x\). Multiply the denominator \(x^2 + 2\) by \(-2x\) to get \(-2x^3-4x\). Subtract this from the new polynomial:
\[

$$\begin{align*} &(-2x^3 - 6x^2 - x - 4)-(-2x^3 - 4x)\\ =&-2x^3 - 6x^2 - x - 4 + 2x^3 + 4x\\ =& - 6x^2 + 3x - 4 \end{align*}$$

\]

Step3: Divide the new leading term

Divide the leading term of the new polynomial \(-6x^2\) by \(x^2\), we get \(-6\). Multiply the denominator \(x^2 + 2\) by \(-6\) to get \(-6x^2-12\). Subtract this from the new polynomial:
\[

$$\begin{align*} &(-6x^2 + 3x - 4)-(-6x^2 - 12)\\ =&-6x^2 + 3x - 4 + 6x^2 + 12\\ =& 3x + 8 \end{align*}$$

\]

Step4: Combine the results

The quotient terms we got are \(3x^2\), \(-2x\), \(-6\) and the remainder is \(3x + 8\). So the division result is \(3x^2-2x - 6+\frac{3x + 8}{x^2 + 2}\)

Answer:

C. \(3x^2 - 2x - 6 + \frac{3x + 8}{x^2 + 2}\)