Prove or disprove: $ \frac{3x^3+2x+1}{x+2} $ is $ \theta (x^2) $.

Question

Prove or disprove: $ \frac{3x^3+2x+1}{x+2} $ is $ \theta (x^2) $.

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I know that $f(x)$ is $\theta (g(x)) $ if it is both $ O(g(x)) $ and $\Omega (g(x))$ when $ x > n$

I reasoned that $f(x)$ is $O(x^2)$ when $C= 6$ and $ x> 1$

I also reasoned that $f(x) $ is $\Omega(x^2)$ if $C=1$ and $x>0$.

I am not sure if I'm correct though...

Any help?

Answer 1

Looks right to me. I'd make it obvious that $x^2$ is the highest order term: turn it into polynomial-plus-remainder form (which may have a real name), using long or synthetic division. This gives

$$\frac{3x^3+2x+1}{x+2}=3x^2-6x+14-\frac{27}{x+2}$$

Which makes life a lot easier: $3x^2$ obviously dominates for large $x$. Pick some $C$ values above and below $3$ and you win.