what is the quotient of $(x^{3}+8)div(x + 2)$?\n$x^{2}+2x + 4$\n$x^{2}-2x + 4$\n$x^{2}+4$\n$x^{2}-4$

Explanation:

Step1: Recall sum - of - cubes formula

We know that $a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2})$. Here, $x^{3}+8=x^{3}+2^{3}$, so $a=x$ and $b = 2$, then $x^{3}+8=(x + 2)(x^{2}-2x + 4)$.

Step2: Perform polynomial division

We want to find $\frac{x^{3}+8}{x + 2}$. Since $x^{3}+8=(x + 2)(x^{2}-2x + 4)$, then $\frac{(x + 2)(x^{2}-2x + 4)}{x + 2}=x^{2}-2x + 4$.

Answer:

$x^{2}-2x + 4$