The functions f and g are defined as follows:
$$D = \{x \in \mathbb{R} \mid x \ne -1 \wedge x \ne 1\}$$
$$f: D → \mathbb{R}, f(x) = \frac{4x^{2} + x^{2} - 4x - 1}{x ^{2} - 1}$$
$$g: D → \mathbb{R}, g(x) = \frac{2x^{3} + 3x^{2} - x + 2}{x^{2 } - 1}$$
Represent the functions in the form $$p(x)+ \frac{q(x)}{x^{2} - 1}$$ where p is a first degree polynomial, and q is either 0 or a polynomial whose degree is at most 1.
Does anyone know how to solve this problem? Are there other ways to solve it or is it only polynomial division?
Function f:
$$4x^2+x^2-4x-1=(4x^2-4x-1)+x^2$$
Then I feel that I should factor it further, but don't know how... the denominator of both g and f is just $(x+1)(x-1)$.
Function g:$$2x^3+3x^2-x+2=(2x^ 3-x)+3x^2+2$$
How should I proceed? I'm not the best when it comes to completing the square…
So starting with $4x^3+x^2+4x-1$ we can subtract off $4x(x^2-1)=4x^3-4x$ to get $x^2+8x-1$, or said another way $4x^3+x^2+4x-1 = 4x(x^2-1) + x^2+8x-1$ We can again subtract off $x^2-1$ to get $8x$ or by the same reasoning as before $4x^3+x^2+4x-1 = 4x(x^2-1) + (x^2-1) +8x$.
Now there is no polynomial that I can multiply $x^2-1$ by to get $8x$ so that's our remainder and in general it will be when the degree of the polynomial is strictly lower than the one you're dividing by. This means we're done so dividing our new form by $x^2-1$ we can see that $$\frac{4x^3+x^2+4x-1}{x^2-1}=\frac{4x(x^2-1) + (x^2-1) +8x}{x^2-1}=4x+1+\frac{8x}{x^2-1}$$ and we have the desired form. Note that all we did here was long division in the same manner that we would use for the integers. The other polynomial is solved in the same manner.