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}\\) a. \\(3x^2 - 2x - 5 + \frac{3x + 6}{x^2 + 2}\\) b. \\(3x^2 + 2x + \frac{3x - 4}{x^2 + 2}\\) c. \\(3x^2 - 2x - 6 + \frac{3x + 8}{x^2 + 2}\\) d. \\(3x^2 + 2x - \frac{5x - 8}{x^2 + 2}\\)

Explanation:

Step1: Divide leading terms

Divide $3x^4$ by $x^2$: $\frac{3x^4}{x^2}=3x^2$
Multiply divisor $x^2+2$ by $3x^2$: $3x^2(x^2+2)=3x^4+6x^2$
Subtract from dividend:
$$(3x^4-2x^3-x-4)-(3x^4+6x^2)=-2x^3-6x^2-x-4$$

Step2: Divide new leading terms

Divide $-2x^3$ by $x^2$: $\frac{-2x^3}{x^2}=-2x$
Multiply divisor by $-2x$: $-2x(x^2+2)=-2x^3-4x$
Subtract from current remainder:
$$(-2x^3-6x^2-x-4)-(-2x^3-4x)=-6x^2+3x-4$$

Step3: Divide next leading terms

Divide $-6x^2$ by $x^2$: $\frac{-6x^2}{x^2}=-6$
Multiply divisor by $-6$: $-6(x^2+2)=-6x^2-12$
Subtract from current remainder:
$$(-6x^2+3x-4)-(-6x^2-12)=3x+8$$

Step4: Combine quotient and remainder

The quotient polynomial is $3x^2-2x-6$, with remainder $3x+8$. Write as:
$3x^2-2x-6+\frac{3x+8}{x^2+2}$

Answer:

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