Answer:
Hello,
Step-by-step explanation:
[tex]\dfrac{f(x)}{3} =\dfrac{x^4+x^3+x^2+1}{(x-1)(x+2)} \\\\=\dfrac{(x^2+3)(x-1)(x+2)-3x+7}{(x-1)(x+2)} \\=x^2+3-\dfrac{3x-7}{(x-1)(x+2)} \\\\=x^2+3-\dfrac{3}{x-1} +\dfrac{1}{(x-1)(x-2)} \\\\\dfrac{f(x)}{3}-\dfrac{3x^2+9}{3} =-\dfrac{3}{x-1} +\dfrac{1}{(x-1)(x-2)} \\\\\\ \lim_{x \to +\infty} (\dfrac{f(x)}{3}-\dfrac{3x^2+9}{3} )\\\\=0+0=0\\\\\\P(x)=-x^2-3[/tex]
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}[/tex
So 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]