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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
avatar+2248 
+3

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
avatar+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
avatar+1183 
+1

Well X2, you discovered your own sin, repented and made reparations. You’re forgiven and absolved.laugh

GingerAle  Apr 11, 2018

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