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 $y=0$ as $x$ approaches positive infinity.

2 answers

Question:

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 $y=0$ as $x$ approaches positive infinity.

Answer: [tex]g(x)=-3x^2-9[/tex]Explanation:[tex]3\frac{x^4+x^3+x^2+1}{x^2+x-2}[/tex]+[tex]\frac{p(x)(x^2+x-2)}{x^2+x-2}[/tex]We need p(x) need to be a degree 2 polynomial so the numerator of the second fraction is degree 4. Our goal is to cancel the terms of the first fraction's numerator that are of degree 2 or higher.So let p(x)=ax^2+bx+c.[tex]3\frac{x^4+x^3+x^2+1}{x^2+x-2}[/tex]+[tex]\frac{p(x)(x^2+x-2)}{x^2+x-2}[/tex]Plug in our p:[tex]3\frac{x^4+x^3+x^2+1}{x^2+x-2}[/tex]+[tex]\frac{(ax^2+bx+c)(x^2+x-2)}{x^2+x-2}[/tex]Take a break to multiply the factors of our second fraction's numerator.Multiply:[tex](ax^2+bx+c)(x^2+x-2)[/tex]=[tex]ax^4+ax^3-2ax^2[/tex]+[tex]bx^3+bx^2-2bx[/tex]+[tex]cx^2+cx-2c[/tex]=[tex]ax^4+(a+b)x^3+(-2a+b+c)x^2+(-2b+c)-2c[/tex]Let's go back to the problem:[tex]3\frac{x^4+x^3+x^2+1}{x^2+x-2}[/tex]+[tex]\frac{ax^4+(a+b)x^3+(-2a+b+c)x^2+(-2b+c)x-2c}{x^2+x-2}[/tex]Let's distribute that 3:[tex]\frac{3x^4+3x^3+2x^2+3}{x^2+x-2}[/tex]+[tex]\frac{ax^4+(a+b)x^3+(-2a+b+c)x^2+(-2b+c)x-2c}{x^2+x-2}[/texSo this forces [tex]a=-3[/tex].Next we have [tex]a+b=-3[/tex]. Based on previous statement this forces [tex]b=0[/tex].Next we have [tex]-2a+b+c=-3[/tex]. With [tex]b=0[/tex] and [tex]a=-3[/tex], this gives [tex]6+0+c=-3[/tex].So [tex]c=-9[tex].Next we have the x term which we don't care about zeroing out, but we have [tex]-2b+c[/tex] which equals [tex]-2(0)+-9=-9[/tex].Lastly, [tex]-2c=-2(-9)=18[/tex].This makes [tex]p(x)=-3x^2-9[/tex].This implies [tex]g(x)\frac{(-3x^2-9)(x^2+x-2)}{x^2+x-2}[/tex] or simplified [tex]g(x)=-3x^2-9[/tex]

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