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**+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**+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