What would the steps be to solve this limit? We are currently learning how to evaluate limits in my Calculus class.  I'm starting to get  a lot better at them however I got this one wrong.  I know the answer is 12 but I'm not sure how  to get it.  The limit is as x approaches -2, of (x^3 + 8)/(x+2).  I'm guessing you need to factor out the numerator.  Are there any special tricks to factor out a trinomial?  Can someone show me step by step how to solve this, it would really help me a lot in understand how to do these types of problems, thanks
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First, and to start, it will be good to recall or learn the identities for the following cubics:
$$a^3+b^3=(a+b)(a^2-ab+b^2)\;\;\text{and}\;\;a^3-b^3=(a-b)(a^{2}+ab+b^2)$$
This, in fact generalizes, and whatever the positive integer $n$, when you have $a^n + b^n$, $(a + n)$ is always a factor. When you have $a^n - b^n$, $(a - b)$ is always a factor.
In your case, given $(x^3 + 8) = (x^3 + 2^3)$, that means that $(x + 2)$ is a factor. Using polynomial division, or the first identity listed above, we can find the remaining factor: $(x^3 + 8) = x^3 + 2^3 = (x + 2)(x^2 - 2x + 4)$. This gives us:
$$\lim_{x \to -2} \frac{x^3 + 8}{x+2} \quad = \quad \lim_{x \to -2} \frac{(\cancel{x + 2})(x^2 - 2x + 4)}{(\cancel{x + 2})} \quad \overset{x\neq -2}  = \quad\lim_{x\to -2} x^2 - 2x + 4 = 12$$
A: Hint: $$\frac{x^3+8}{x+2} = \frac{x^3+2^3}{x+2}= \frac{(x+2)(x^2-2x+2^2)}{x+2}= x^2-2x+4.$$
A: Since you knew that $x=-2$ resulted in $x^3+8=0$, doesn't that mean $x=-2$ is a root of $x^3+8$? Why not try polynomial division with $x+2$ since you now know it's a factor?

Despite the fact you could easily do this problem with the common identity $a^3+b^3=(a+b)(a^2-ab+b^2)$ and in general $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\dots+b^{n-1})$, the above logic is far more general and will help you solve a greater class of problems.
