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As above, let $$f(x) = 3\cdot\frac{x^4+x^3+x^2+1}{x^2+x-2}.$$Give a polynomial $g(x)$ so that $f(x) + g(x)$ has a horizontal asymptote of 0

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Let $$f(x) = 3\cdot\frac{x^4+x^3+x^2+1}{x^2+x-2}.$$ Give a polynomial g(x) so that f(x) + g(x) has a horizontal asymptote of 0 as x approaches positive infinity.

used latex on this one

Guest Apr 10, 2018
#1
+2248
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Asymptotes are always expressed as equations because they are lines, so I would assume that the horizontal asymptote lies at y=0.

Of course, horizontal asymptotes have many rules associated with them. In this case, we are concerned with the following rule: If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptotes exists at the x-axis (also known as y=0).

If we take $$f(x) = \frac{3(x^4+x^3+x^2+1)}{x^2+x-2}$$ and multiply it by $$\frac{1}{x^3}$$, then we would have finished this problem. Therefore, let's find g(x) that does this.

 $$\frac{3(x^4+x^3+x^2+1)}{x^2+x-2}+g(x)=\frac{3(x^4+x^3+x^2+1)}{x^3(x^2+x-2)}$$ Let's do some subtraction here. $$g(x)=\frac{3(x^4+x^3+x^2+1)}{x^3(x^2+x-2)}-\frac{3(x^4+x^3+x^2+1)}{(x^2+x-2)}$$ Let's convert the rightmost fraction into one with a common denominator so that further simplification is possible. $$g(x)=\frac{3(x^4+x^3+x^2+1)}{x^3(x^2+x-2)}-\frac{3x^3(x^4+x^3+x^2+1)}{x^3(x^2+x-2)}$$ Let's do some distributing. $$g(x)=\frac{3x^4+3x^3+3x^2+3}{x^3(x^2+x-2)}-\frac{(3x^7+3x^6+3x^5+3x^3)}{x^3(x^2+x-2)}$$ Now, subtract! $$g(x)=\frac{-3x^7-3x^6-3x^5+3x^4+3x^2+3}{x^3(x^2+x-2)}\\$$ Not a lot of simplification happens, though, but let's do it anyway!

You should see that, when adding g(x) to f(x), this results in a horizontal asymptote of y=0.

TheXSquaredFactor  Apr 11, 2018
#2
+2248
+3

Well, I guess I committed a cardinal sin: I did not factor completely. I thought there was no way that either the numerator or denominator could have a common factor. How wrong I was!

 $$-3x^7-3x^6-3x^5+3x^4+3x^2+3$$ First, factor out a common factor of -3 from every term. $$-3(x^7+x^6+x^5-x^4-x^2-1)$$ Well, I can use the rational root theorem to determine that 1 is a root of the seven-degree polynomial, so $$x-1$$ must also be a factor. This results in the following result. After using synthetic division, I got the following! $$-3\left(x-1\right)\left(x^6+2x^5+3x^4+2x^3+2x^2+x+1\right)$$ Let's worry about the denominator. $$x^3(x^2+x-2)$$ I was able to use factoring here, too, by finding the products of -2 that equal 1: 2 and -1. Let's complete the factoring, then! $$x^3(x-1)(x+2)$$ $$\frac{-3\left(x-1\right)\left(x^6+2x^5+3x^4+2x^3+2x^2+x+1\right)}{x^3(x-1)(x+2)}$$ Look at that! $$x-1$$ is a common factor. That's crazy! This leaves us with the following! $$g(x)=\frac{-3\left(x^6+2x^5+3x^4+2x^3+2x^2+x+1\right)}{x^3(x+2)}, x(x+2)\neq0\text{ and }x\neq 1$$ I guess this is better!
TheXSquaredFactor  Apr 11, 2018
#3
+1183
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Well X2, you discovered your own sin, repented and made reparations. You’re forgiven and absolved.

GingerAle  Apr 11, 2018