Question

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

Explanation:

Step1: Divide the leading terms

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: \((3x^4 - 2x^3 - x - 4)-(3x^4 + 6x^2)=-2x^3-6x^2 - x - 4\).

Step2: Divide the new leading term

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

Step3: Divide the new leading term

Divide the leading term of \(-6x^2 + 3x - 4\) (which is \(-6x^2\)) by the leading term of the denominator \(x^2\), we get \(-6\). Multiply the denominator \(x^2 + 2\) by \(-6\) to get \(-6x^2-12\). Subtract this from \(-6x^2 + 3x - 4\): \((-6x^2 + 3x - 4)-(-6x^2-12)=3x + 8\).

Step4: Combine the results

The quotient terms are \(3x^2-2x - 6\) and the remainder is \(3x + 8\), so the expression is \(3x^2-2x - 6+\frac{3x + 8}{x^2+2}\).

Answer:

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