Where does the x come from when you decompose non-linear denominators?

Question

I have done >4h of research with different text books and I'm still stucked. I also have looked at enter link description here

But still I don't get it, where does this x at Bx+C come from?:

$$ \frac{3}{(x+1)(x²+4)} = \frac{A}{(x+1)}+\frac{Bx+C}{(x²+4)} $$

The thing I am struggling with is: I don't have any clue about the connection between the denominator and the numerator on the right hand side. All I know is that the numerators of the right hand side should give, when multiplied by the denominators 3 and that the product of the denominators should be (x+1)(x²+4). So how do we know that there has to be an x in Bx+C ?

My textbook says: "Now if we are dividing by x²+4 the remainder can have xs in, as well as numbers" And I don't know what "we" are dividing to get a remainder like that ? 3/(x²+4) ?

Maybe someone could give a really really really easy step by step example, how you receive that Bx term rather than why it is logical to be there?

Like my textbook mentioned, there's a way to simply divide "something" and receive a remainder of that form .

Answer 1

As mentioned in one of my comments, Hagen von Eitzen's answer to the linked question may provide a satisfactory answer.

If you want to see an explicit computation, consider a modification of your example: $$ \frac{10x^2+5}{(x+1)(x^2+4)}. $$ Noting that $x+1$ and $x^2+4$ have no non-constant polynomial factor in common, we can, using the Euclidean algorithm, write $1=u(x)(x^2+4)+v(x)(x+1)$. To find $u(x)$ and $v(x)$ note that $x^2+4$ divided by $x+1$ gives quotient $x-1$ and remainder $5$. So $5=(x^2+4)-(x-1)(x+1)$, implying that $u(x)=\frac{1}{5}$ and $v(x)=-\frac{1}{5}(x-1)$. We may now write the original rational function as $$ \begin{aligned} \frac{(2x^2+1)\cdot5}{(x+1)(x^2+4)}&=\frac{(2x^2+1)[(x^2+4)-(x-1)(x+1)]}{(x+1)(x^2+4)}=\frac{2x^2+1}{x+1}-\frac{(2x^2+1)(x-1)}{x^2+4}\\ &=\frac{2x^2+1}{x+1}-\frac{2x^3-2x^2+x-1}{x^2+4}. \end{aligned} $$ By polynomial division we may write both terms in the form $$ \frac{\text{numerator}}{\text{denominator}}=\text{quotient}+\frac{\text{remainder}}{\text{denominator}} $$ Since the numerator of the original rational function has lower degree than does the denominator, it is clear that the quotient terms are going to have to cancel, leaving only the remainder terms. Indeed $$ \frac{2x^2+1}{x+1}=2x-2+\frac{3}{x+1} $$ and $$ \frac{2x^3-2x^2+x-1}{x^2+4}=2x-2+\frac{-7x+7}{x^2+4}, $$ leaving $$ \frac{10x^2+5}{(x+1)(x^2+4)}=\frac{3}{x+1}+\frac{7x-7}{x^2+4}. $$

Answer 2

Technically there is a way to go through this just in terms of algebra, but I think it is easier to just talk about an example. The simplest case where this comes up that I can think of is:

$$\frac{1}{x(x^2+1)}=\frac{1}{x}-\frac{x}{x^2+1}.$$

You might guess $\frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{B}{x^2+1}$. One clue that this can't be right is that it has the wrong scaling: the left side behaves like $x^{-3}$ as $x \to \infty$ while the right side can only behaves as $x^{-1}$ as $x \to \infty$. There is no way the second term can "cancel out enough" of the first term to give the $x^{-3}$ scaling.

But let's say you don't know anything about scaling and just want to think about coefficients. You clear denominators and get

$$1=A(x^2+1)+Bx$$

How will the $x^2$ and $x$ terms be eliminated? There is no way, short of $A=B=0$ which is also incorrect. But if $B$ were instead $Bx+C$, then you would have a $Bx^2$ term there to be able to cancel out the $Ax^2$. Conversely, just $Bx$ would not be sufficient if the numerator were not constant, because then there would be no linear term to balance the linear term on the left side of the equation. In general you need both.

A way to see it as not being a special case is to go to complex numbers. Note that this is usually more trouble than it's worth when actually trying to find antiderivatives of rational functions because of issues with ambiguity of the complex logarithm. But for illustration it helps.

You have:

$$\frac{1}{x(x^2+1)}=\frac{A}{x}+\frac{B}{x-i}+\frac{C}{x+i}$$

for complex constants $A,B,C$. This is the same as the usual non-repeated linear factor case. You could compute $A,B,C$ directly and then simplify, but you can also get a common denominator in the second and third terms while still leaving everything symbolic:

$$\frac{B}{x-i}+\frac{C}{x+i}=\frac{B(x+i)+C(x-i)}{x^2+1}=\frac{(B+C)x+i(B-C)}{x^2+1}.$$

Thus $\frac{B}{x-i}+\frac{C}{x+i}=\frac{Dx+E}{x^2+1}$ where $D=B+C,E=i(B-C)$. Note that a linear term appears when you try to force two terms with real roots to have a common denominator, for exactly the same algebraic reason. For example:

$$\frac{1}{x-1}+\frac{1}{x+2}=\frac{x+2+x-1}{(x-1)(x+2)}=\frac{2x+1}{(x-1)(x+2)}.$$

From this perspective the linear term shows up because in choosing to do the decomposition over $\mathbb{R}$ we are forced to combine two terms in the decomposition over $\mathbb{C}$ that have different roots.