SOLUTION: Give a polynomial g(x) so that f(x) + g(x) has a horizontal asymptote of y = 0 as x approaches positive infinity, where f(x) = \frac{2x^4 - 3x^3 - 8x^2 + 4x

SOLUTION: Give a polynomial g(x) so that f(x) + g(x) has a horizontal asymptote of y = 0 as x approaches positive infinity, where f(x) = \frac{2x^4 - 3x^3 - 8x^2 + 4x - 4}{x^2 + x}.

Question 1209878: Give a polynomial g(x) so that f(x) + g(x) has a horizontal asymptote of y = 0 as x approaches positive infinity, where
f(x) = \frac{2x^4 - 3x^3 - 8x^2 + 4x - 4}{x^2 + x}.

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let $f(x) = \frac{2x^4 - 3x^3 - 8x^2 + 4x - 4}{x^2 + x}$. We want to find a polynomial $g(x)$ such that $f(x) + g(x)$ has a horizontal asymptote of $y = 0$ as $x \to \infty$.
First, we perform polynomial long division on $f(x)$:
```
2x^2 - 5x - 3
____________________
x^2+x | 2x^4 - 3x^3 - 8x^2 + 4x - 4
2x^4 + 2x^3
--------------------
-5x^3 - 8x^2
-5x^3 - 5x^2
--------------------
-3x^2 + 4x
-3x^2 - 3x
--------------------
7x - 4
```
So, $f(x) = 2x^2 - 5x - 3 + \frac{7x - 4}{x^2 + x}$.
As $x \to \infty$, the term $\frac{7x - 4}{x^2 + x}$ approaches 0.
Thus, as $x \to \infty$, $f(x)$ behaves like $2x^2 - 5x - 3$.
We want $f(x) + g(x)$ to have a horizontal asymptote of $y = 0$. This means that as $x \to \infty$, $f(x) + g(x)$ should approach 0.
So, we need $g(x)$ to cancel out the polynomial part of $f(x)$, which is $2x^2 - 5x - 3$.
Therefore, we can choose $g(x) = -(2x^2 - 5x - 3) = -2x^2 + 5x + 3$.
Then, $f(x) + g(x) = 2x^2 - 5x - 3 + \frac{7x - 4}{x^2 + x} - 2x^2 + 5x + 3 = \frac{7x - 4}{x^2 + x}$.
As $x \to \infty$, $\frac{7x - 4}{x^2 + x} \to 0$.
Thus, $f(x) + g(x)$ has a horizontal asymptote of $y = 0$ as $x \to \infty$.
Final Answer: The final answer is $\boxed{-2x^2 + 5x + 3}$

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