How do I compute the individual terms of a polynomial to the power of -1?

Question

If my polynomial $p$ is: $x+1$, obviously $p^{-1} = \frac{1}{x+1}$. Is it possible for me to split $\frac{1}{x+1}$ into a sum of two terms? In other words, is there an algorithm to write $p^{-1}$ as another polynomial other than just writing $\frac{1}{p}$?

Answer 1

It is not possible to write $1/(x+1)$ as a polynomial in $x$. There are many ways to see this, probably the easiest is that the polynomial would give a value at $x=-1$, but $1/(x+1)$ has no value at $-1$.

Answer 2

If $\quad|x|<1,\quad$ then $\quad\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n.\quad$ Now, exchange x with $-x$.

Answer 3

For your case, it is impossible to write $\dfrac{1}{x+1}$ as a combination of two fractions. But you can with fractions that have denominators with degree $2$ or higher. You do this using partial fraction decomposition. For example, the fraction $$\dfrac{1}{x^2-5x+6}$$ can be "split up" into two terms. What you need to do is factor the denominator, and split the fraction apart. Let me show you what I mean. $$\dfrac{1}{x^2-5x+6}$$ $$=\dfrac{1}{(x-2)(x-3)}$$ $$=\dfrac{A}{x-2}+\dfrac{B}{x-3}$$ You have to assume variables $A$ and $B$ (they are both constants). Before I forget, do not assume that you are always going to be doing partial fractions this way. There are different ways for different kinds of fractions. But for this example, we have all linear, non-different factors. So it is ok to just assume the constants $A$ and $B$. For linear factors, there will be no $x$ in the numerator.

Now we multiply both $\dfrac{A}{x-2}+\dfrac{B}{x-3}$ and $\dfrac{1}{x^2-5x+6}$ by $x^2-5x+6$. This will eliminate all of our denominators. $$\dfrac{A(x-2)(x-3)}{x-2}+\dfrac{B(x-2)(x-3)}{x-3}=\dfrac{1(x^2-5x+6)}{x^2-5x+6}$$ $$A(x-3)+B(x-2)=1$$ We have to solve for $A$ and $B$. To do this, let's pick values of $x$ that will eliminate one of the two. Let's pick $x=3$ to eliminate $A$ first. $$A(3-3)+B(3-2)=1$$ $$1B=1$$ $$B=1$$ I do not think you can just plug the value of $B$ back into the equation to find $A$. Instead, we have to pick a value of $x$ to eliminate $B$. This time, we will pick $x=2$. $$A(2-3)+B(2-2)=1$$ $$A(-1)=1$$ $$A=-1$$ Final answer: $$\dfrac{1}{x^2-5x+6}=\dfrac{-1}{x-2}+\dfrac{1}{x-3}$$ That can be rewritten as: $$\boxed{\dfrac{1}{x^2-5x+6}=\dfrac{1}{x-3}-\dfrac{1}{x-2}}$$ Conclusion: Yes it is possible to write rational expressions as a combination of 2 or more fractions with partial fraction decomposition.

If you want to learn more about partial fractions, look it up on Google. I hope my answer helped you.

More information about partial fractions if you want to know:

Partial fraction decomposition becomes very important later on in calculus. To start, let's decompose $\dfrac{3x+20}{x^2+4x}$ into partial fractions. $$\dfrac{3x+20}{x^2+4x}$$ $$=\dfrac{3x+20}{x(x+4)}$$ Split into two fractions with constants $A$ and $B$ as the numerators: $$\dfrac{A}{x}+\dfrac{B}{x+4}=\dfrac{3x+20}{x^2+4x}$$ Get rid of the denominators... $$\dfrac{A(x)(x+4)}{x}+\dfrac{B(x)(x+4)}{x+4}=\dfrac{(3x+20)(x^2+4x)}{x^2+4x}$$ $$A(x+4)+Bx=3x+20$$ This time we will not use the number-substitution to eliminate the variables one at a time, as that method doesn't always work. Instead we will expand the left hand side of the equation. $$A(x+4)+Bx=3x+20$$ $$Ax+4A+Bx=3x+20$$ Now, we will do something called equating the coefficients. This means that because $Ax+Bx=3x$, then $A+B=3$. Likewise, $4A=20$ (because both do not have an $x$ attached to it). We can easily find out that $A=5$ from simple algebra. Now we will substitute the value of $A$ into $A+B=3$ to solve for $B$. $$5+B=3$$ $$B=-2$$ The partial fraction decomposition for $\dfrac{3x+20}{x^2+4x}$ is: $$\boxed{\dfrac{3x+20}{x^2+4x}=\dfrac{5}{x}-\dfrac{2}{x+4}}$$ When integrating this rational expression, it is much easier to integrate $\dfrac{5}{x}-\dfrac{2}{x+4} \ dx$ than it is to integrate $\dfrac{3x+20}{x^2+4x} \ dx$. It is possible to integrate the latter by itself, but it requires many steps, and can get complicated. Let me show you how to integrate the rational expression. $$\int \dfrac{3x+30}{x^2+4x} \ dx$$ $$=\int \dfrac{5}{x}-\dfrac{2}{x+4} \ dx$$ $$=\int \dfrac{5}{x} \ dx - \int \dfrac{2}{x+4} \ dx$$ $$=5\int \dfrac{1}{x} \ dx - 2\int \dfrac{1}{x+4} \ dx$$ $$=5\ln (x) - 2\ln (x+4) + C$$ Final Answer: $$\boxed{\int \dfrac{3x+20}{x^2+4x} = 5\ln (x) - 2\ln (x+4) + C}$$