Basic functions: Rational functions
Partial fraction decomposition: irreducible quadratic factors for the denominator
Not every quadratic polynomial can be factorized into linear factors when only real coefficients are allowed: a simple example is the polynomial \(x^2+1\), which has a negative discriminant. We first look at two examples of rational functions with a denominator that has irreducible quadratic factors.
Example 1 We consider the rational function \(\frac{3}{x^3+1}\), whose denominator can be fully factorised as \((x+1)(x^2-x+1)\) (check this!). In partial fraction decomposition we try to rewrite this fraction as follows: \[\frac{3}{x^3+1}=\frac{\mathrm{something}}{x+1}+ \frac{\mathrm{SOMETHING}}{x^2-x+1}\] So we need to find \(\mathrm{something}\) and \(\mathrm{SOMETHING}\) such that \[\mathrm{something}\cdot(x^2-x+1)+ \mathrm{SOMETHING}\cdot(x+1)=3\] We cannot use just real numbers for this. Next we try to achieve this with polynomials and, after some thought, we can reasonably try the following: \[a(x^2-x+1)+ (bx+c)(x+1)=3\] where \(a\), \(b\) and \(c\) are still unknown numbers. Eliminating brackets in the right-hand side and collecting like powers of \(x\) gives \[(a+b)x^2+(b+c-a)x+a+c=2\] This must be valid for all \(x\) and therefore we have the following three linear equations: \[a+b=0,\quad b+c-a=0,\quad a+c=3\text.\] The solution of this system of equations is \[a=1,\quad b=-1,\quad c=2\text.\] The partial fraction decomposition is therefore \[\frac{3}{x^3+1}=\frac{1}{x+1} + \frac{-x+2}{x^2-x+1}\text.\]
Example 2 Consider the rational function \(\frac{4x^2+3}{9x^5+6x^3+x}\), whose denominator can be fully factorised as \(x(3x^2+1)^2\) (check this!). By analogy with the method with repeated linear factors, we try the following decomposition: \[\frac{4x^2+3}{9x^5+6x^3+x}=\frac{a}{x}+\frac{bx+c}{3x^2+1}+\frac{dx+e}{(3x^2+1)^2}\] for yet unknown numbers \(a\), \(b\), \(c\), \(d\) and \(e\). \[\begin{aligned}4x^2+3 &=a(3x^2+1)^2+x(3x^2+1)(bx+c)+x(dx+e)\\[0.25cm] &=a(9x^4+6x^2+1)+x(3bx^3+3cx^2+bx+c)+ dx^2+ex\\[0.25cm]
&=(9a+3b)x^4+3cx^3+(6a+b+d)x^2+(c+e)x+a\end{aligned}\] must then be valid for all values of \(x\) . So we get a system of 5 linear equations in 5 unknowns: \[9a+3b=0,\quad 3c=0,\quad 6a+b+d=4,\quad c+e=0,\quad a=3\text.\] The solution of this system can be found after some puzzling and is equal to \[a=3,\quad b=-9,\quad c=4,\quad c=0,\quad d=-5\quad e=0\text.\] The partial fraction decomposition is therefore \[\frac{4x^2+3}{9x^5+6x^3+x}=\frac{3}{x}-\frac{9x}{3x^2+1}-\frac{5x}{(3x^2+1)^2}\]
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The above examples illustrate how patrial fraction decomposition can be done for a rational function \(\frac{p(x)}{q(x)}\) with the degree of \(p(x)\) less than the degree of \(q(x)\) and where the factor decomposition of \(q\) consists of irreducible quadratic factors and linear factors, possibly with repetitions. Every distinct factor may therefore be raised to a certain power with positive natural exponent in the factorisation of the denominator. This brings us to the general case of partial fraction decomposition of a rational function!
Partial fraction decomposition for a denominator that can be factorised into linear and irreducible quadratic factors, possibly with repetitions We consider a rational function \(\frac{p(x)}{q(x)}\) with the degree of \(p(x)\) less than the degree of \(q(x)\) and with \(q(x)\) a polynomial that can be written as \(n\) distinct linear factors and \(s\) distinct irreducible quadratic factors say, \[\begin{aligned}q(x)&=(x-\alpha_1)^{n_1}\cdot (x-\alpha_2)^{m_1}\cdots(x-\alpha_n)^{m_n}\\[0.25cm]
&\phantom{=}{}\cdot(x^2+\beta_1x+\gamma_1)^{r_1}\cdot (x^2+\beta_2x+\gamma_2)^{r_2}\cdots(x^2+\beta_sx+\gamma_s)^{r_s}\end{aligned}\] \(a_i\neq\alpha_j\) for \(i\neq j, 1\le i,j\le n\) and \(m_i\in\{1,2,3,\ldots\}\) for \(1\le i\le n\), and with \((b_i,\gamma_i)\neq(b_j,\gamma_j)\) for \(i\neq j, 1\le i,j,\le s\) and \(r_i\in\{1,2,3,\ldots\}\) for \(1\le i\le s\). Then, each power of a linear factor, say \((x-\alpha)^m\), in the partial fraction decomposition of \(\frac{p(x)}{q(x)}\) corresponds with the following sum of fractions: \[\frac{a_1}{x-\alpha}+\frac{a_2}{(x-\alpha)^2}+\cdots +\frac{a_m}{(x-\alpha)^m}\] Each power of an irreducible quadratic factor, say \((x^2+\beta x+\gamma)^r\), corresponds with the following sum of fractions: \[\frac{b_1x+c_1}{x^2+\beta x+\gamma}+\frac{b_2x+c_2}{(x^2+\beta x+\gamma)^2}+\cdots +\frac{b_rx+c_r}{(x^2+\beta x+\gamma)^r}\] All coefficients can be computed by adding up all fractions in the decomposition and equating the coefficients of like powers of \(x\) in the numerator with those in \(p(x)\), followed by solving the obtained system of equations.
