Derivation of the general forms of partial fractions
Solution 1:
Your first and second cases are equivalent. You need to understand complex numbers to be able to see that. The complex numbers, written $\mathbb{C}$ are of the form $x+iy$ where $x$ and $y$ are real numbers that you already know and $i$ is a special number with $i^2 = -1$. Usually, we have $i = \sqrt{-1}$. To see that the second case is the same as the first, notice that:
$$\frac{2x+3}{(x-1)(x^2+4)} \equiv \frac{A}{x-1}+\frac{B}{x-2i}+\frac{C}{x+2i}$$
where $A=1$, $B=-\frac{1}{2}-i\frac{1}{4}$ and $C=-\frac{1}{2}+i\frac{1}{4}$. The only reason you are taught the second form is that you don't know about $x\pm 2i$ and they recombine them:
$$\frac{B}{x-2i} + \frac{C}{x+2i} \equiv \frac{1-x}{x^2+4}$$
The third case isn't quite the same. But notice that the denominator is a cubic. When you multiply it out, you'll have an $x^3$ and lower powers. We couldn't get away with just two fraction because
$$\frac{A}{x-a}+\frac{B}{x-b} \equiv \frac{A(x-b)+B(x-a)}{(x-a)(x-b)}$$
and this only has a quadratic expression as the denominator.
Solution 2:
As requested, I'm distilling and simplifying the information from this link: http://caicedoteaching.wordpress.com/2008/11/18/175-partial-fractions-decomposition-2/
First, we need a concept. An irreducible polynomial is one that can't be reduced as a product of smaller-degree polynomials. These are the building blocks of all polynomials and, as you can see with partial fractions, we can usually break the study of composite polynomials into irreducible ones. We also need some facts. First, any polynomial of odd degree has a root. Intuitively, given such a polynomial $p(x)$, if $x$ is large then we can just ignore the lower order terms, so $p$ takes on some positive value, and if $x$ is very negative then the same thing happens but it takes on a negative value (if $p$ had even degree the negative sign would cancel, giving us nothing). So $p$ has to pass through zero somewhere. Second, we can write any quadratic polynomial as the sum of a square and a real number; for instance, $x^2 + 2x + 2 = (x+1)^2 + 1$. This is just completing the square. Third, any polynomial can be written as the product of linear factors and irreducible quadratic factors. This is not a trivial result, and I'm going to assume that you've seen it in the form of the Fundamental Theorem of Algebra, but I'm not going to prove it. The interpretation here is that $q$ is the denominator polynomial we're looking for, and we can really just reduce to the case where $q$ is a product of linear and quadratic factors. The case where these are just linear factors corresponds to your part 1, where the case with quadratic factors is part of your other parts. The "repeated factors" case will fall out of this naturally.
So, given some polynomial $q$ we can write it as $q(x) = C \cdot \prod_{i=1}^k(x-a_i)^{r_i} \cdot \prod_{i=1}^l ((x+b_i) + c_i)^{s_i}$. Then given any polynomial $p$ we can write $\frac{p(x)}{q(x)} = \sum_{i=1}^k \left( \sum_{j=1}^{r_i} \frac{a_{i,j}}{(x-a_i)^{r_i}} \right) + \sum_{i=1}^l \left( \sum_{j=1}^{s_i} \frac{b_{i,j}x + c_{i,j}}{((x+b_i)^2+c_i)^j} \right)$. I'm abusing notation a little, but know that the double-subscripted $a_{i,j}, b_{i,j},c_{i,j}$ are the constants we're looking for in the method of partial fractions, while the single-subscripted ones are the input.
This looks daunting, but it's exactly what you want to show. If $q(x) = (x-1)^2(x^2 + 2x + 2)$ then all we're saying is that, for example, there are the right constants to make the equality we want, $\frac{1}{(x-1)^2(x^2+2x+2)} = \frac{a}{x-1} + \frac{b}{(x-1)^2} + \frac{c}{x^2 + 2x + 2}$, hold. The interpretation in that awful sum is that the inner sums are going over the powers of each factor, i.e. how many times the factors appear in $q$, while the outer ones are going over the factors themselves.
We need a little preparation. First, if $p_1$ and $p_2$ are polynomials and for every $x$ we have $p_1(x) = p_2(x)$, then the coefficients of the polynomials are equal. This is "obvious" and I won't go into it unless requested. Second, if we have polynomials $p_1(x), p_2(x)$, with $p_2(x)$ not just the zero poylnomial, then there are unique polynomials $Q(x)$ and $R(x)$ with $p_1(x) = Q(x)p_2(x) + R(x)$. The $R$ stands for remainder and the $Q$ for quotient; this is just polynomial long division. If you haven't seen it I can explain, but for now I'll leave it be. Third, we need the notion of a greatest common divisor. If $p_1$ and $p_2$ are two polynomials then we can take the factors they have in common and throw them into another polynomial $D$; for instance, if $p_1(x) = (x-1)^2(x^2 + 2x + 2)(x-3)$ and $p_2(x) = (x-1)^2(x^2 + 1)(x^2 + 2)$ then $D = (x-1)^2$. Then $D$ is called the greatest common divisor of $p_1$ and $p_2$, or $\gcd(p_1,p_2) = D$.
Four, a theorem. If $p_1, p_2$ are polynomials and $D$ is their $\gcd$, then there are polynomials $q_1, q_2$ with $p_1Q_1 + p_2q_2 = D$. That is, we can multiply and combine polynomials to get their greatest common divisor. This is known as the Euclidean Algorithm. To prove this, we'll really just apply long division many times. We can assume that the degree of $p_1$ is greater than that of $p_2$, renaming them if necessary. Polynomial long division gives us $Q_0$ and $R_0$ with $p_1 = Q_0p_2 + R$. If $R_0 = 0$ then we can stop; $p_2$ actually divides $p_1$ evenly. If it's not zero we can continue, finding $Q_1, R_1$ with $p_2 = Q_1R_0 + R_1$. Each time, we divide the dividend of the previous step by the remainder of the previous step to get a new remainder. We have to stop eventually, because each time the degree of the dividend goes down at least by 1. When it does stop you end up with, say, the step right before you finish being $d = fg + h$ and the last step being $f = ih$, and for illustration the third-to-last step as $b = de + f$. Then we can use these equations to get $h = d - fg = d - (b - de)g = \cdots$. Eventually this will get you back to an equation like $h = Ap_1 + Bp_2$. But by the last step we had, $h$ divides $f$, and then $h$ divides $fg$ so by the second-to-last step $h$ divides $d$, and then by the same argument it divides $b$, and $\cdots$. Eventually you get that it divides $p_2$ and then $p_1$ by the first two steps. But if $h$ divides both of those then it certainly must divide their greatest common divisor as well, so $h$ divides $D$. Now if $h$ is missing some factors, we can just multiply the equation $h = Ap_1 + Bp_1$ by those missing factors to get what we want. If it's not missing some factors then that's the equation we're looking for.
We're getting close to the result. Let's write $q = d_1\ldots d_k$ where the $d_i$ are irreducible and have no common factors. Basically, we're getting those linear and quadratic terms I talked about early on. If $p$ is any polynomial, we should be able to write $\frac{p}{q} = a_0 + \sum_i \frac{b_i}{d_i}$ where the $b_i$ are some polynomials and $a_0$ is too, but with the degree of $b_i$ necessarily less than the degree of $d_i$. This sort of decomposition has to be true if we want partial fractions to work. We'll argue by induction on the number of irreducible factors of $q$. If $q = d_1$ then we can just use polynomial long division to get a $b_1$ with $p = a_0d_1 + b_0$, or rather $\frac{p}{q} = a_0 + \frac{b_0}{d_1}$. Now we'll move to the induction step. Let's assume that the result is true for $q$ a product of $m - 1$ polynomials that have no common factors. Let $q = d_1\ldots d_m$. Let's define $d = d_2 \ldots d_m$, so the product of all of the factors but the first one. By assumption $d_1$ and $d$ don't have any common factors, so their greatest common divisor is 1, so by the previous paragraph we can find some $e,f$ with $d_1e + df = 1$. Then $\frac{p}{q} = \frac{p\cdot 1}{q} = \frac{ped_1 + pdf}{q} = \frac{ped_1 + pdf}{d_qd} = \frac{pf}{d_1} + \frac{pe}{d}$. But we assumed we proved all the cases $1,\ldots,m-1$ so far, so if we apply the $m=1$ case using $pf$ for $p$ and $d_1$ for $q$ we get $\frac{pf}{d_1} = l_0 + \frac{b_1}{d_1}$ with the degree of $b_1$ less than that of $d_1$. Applying the $m = m-1$ case using $pe$ for $p$ and $d$ for $q$ we get $\frac{pe}{d} = k_0 + \sum_{i=2}^m \frac{b_i}{d_i}$. If you add these you get $\frac{pe}{d} + \frac{pf}{d_1} = (l_0 + k_0) + \sum_i \frac{b_i}{d_i}$ so $\frac{p}{q} = a_0 + \sum \frac{b_i}{d_i}$. We're done with this claim.
Now, can you see how to apply the previous paragraph to prove the result? In this case, $q = \prod (x-a_i){r_i} \prod ((x+b_i)^2+c_i)^{s_i}$ so we can just let each factor $d_i$ be something like $(x-a_i)^{r_i}$ or $((x+b_i)^2+c_i)^{s_i}$. Then we can use the above paragraph to get $\frac{p}{q} = \sum \frac{b_i}{d_i}$. Let's assume for a second that $d_i$ is a power of a linear factor. We can, using polynomial long division, write $b_i = \alpha_0 + \ldots + \alpha_{r_i}(x-a_i)^{r_i}$. Thus when we divide through by $d_i = (x-a_i)^{r_i}$ we get $\frac{b_i}{d_i} = \frac{\alpha_0}{(x-a_i)^{r_i}} + \ldots + \frac{\alpha_{r_i}}{1} = \sum_{j=1}^{r_i} \frac{a_{i,j}}{(x-a_i)^j}$, just as in the statement of the theorem.
Now, if $d_i = ((x + b_i)^2 + c_i)^{s_i}$, the exact same thing happens, except that we don't necessarily have to have a constant in the numerator. When you try to write $b_i = \alpha_0 + \ldots + \alpha_{s_i}((x + b_i)^2 + c_i)^{s_i}$ you're not totally guaranteed that it looks exactly like this. The reason for this is that in polynomial long division, the remainder always has a lower degree than the dividend. In the case of a linear factor this was fine; the only thing lower than linear is just a real number. But in the quadratic case we might end up with a linear remainder, which gives us the $b_{i,j}x + c_{i,j}$ in the theorem statement.
Solution 3:
It comes from two facts for polynomials $f(x)$ and $g(x)$:
We can find $a(x)$ and $b(x)$ such that $\gcd(f(x),g(x))=a(x)f(x)+b(x)g(x)$.
We can write $f(x)=q(x)g(x)+r(x)$, where $\deg(r)<\deg(g)$.
For your second example, we can use fact 1 to write $1=a(x)(x-1)+b(x)(x^2+4)$. Multiplying by $2x+3$ yields: $$2x+3=[(2x+3)a(x)](x-1)+[(2x+3)b(x)](x^2+4)$$
If $\deg((2x+3)a(x))\ge \deg(x^2+4)=2$, then we can use fact 2 to write $(2x+3)a(x)=q(x)(x^2+4)+r(x)$, where $\deg(r)<2$. Then: $$2x+3=r(x)(x-1)+[q(x)(x-1)+(2x+3)b(x)](x^2+4)$$ Comparing degrees, we see that: $$2x+3=r(x)(x-1)+s(x)(x^2+4)$$ where $\deg(r)<2$, and $\deg(s)<1$. Then: $$\frac{2x+3}{(x-1)(x^2+4)}=\frac{r(x)(x-1)+s(x)(x^2+4)}{(x-1)(x^2+4)}=\frac{r(x)}{x^2+4}+\frac{s(x)}{x-1}=\frac{Ax+B}{x^2+4}+\frac{C}{x-1}$$
Solution 4:
Assad, remember these are fractions, where the NUM has to be of lesser degree than the DENOM or than any factor of the DENOM. So if the DENOM is a non factorable quadratic, the NUM is allowed to be linear, as you are indicating. If the DENOM is repeating, where the lowest term is linear, as in your example, than all of its factors have to be considered with constants in each NUM. I don't think there is more to it, except that if you add them, you should get your original fractions back. Fraction addition ought to be the previous section of partial fraction decomposition. Once you understand the rudiments of fraction addition, you get a grip on PFD. Other than that, it's a matter of practice. Classify your DENOM, set up the partial fractions' DENOM's and work it out. You may post some examples in seperate posts and we can help you to solve them.