2.3 When the denominator has non-repeated quadratic factors

Question 6. Resolve $\left[ {\frac{{7{x^2} - 25x + 20}}{{2(x - 4)({x^2} + 3x + 4)}}} \right]$ into partial fractions.

Solution: As the denominator has non-repeated quadratic factors.

The quadratic polynomial ${({x^2} + 3x + 4)}$ cannot be factorised into linear factors as, $$\begin{equation} \begin{aligned} \Rightarrow D = {b^2} - 4ac = 9 - 4(1)(4) = - 7 \\ \therefore D < 0 \\\end{aligned} \end{equation} $$ It states that the quadratic polynomial ${({x^2} + 3x + 4)}$ has no real solution. Som it cannot be factorised into linear factors.

So, the fraction is expressed as the sum of partial fraction as shown below, $$\begin{equation} \begin{aligned} \Rightarrow \frac{{7{x^2} - 25x + 20}}{{2(x - 4)({x^2} + 3x + 4)}} = \frac{A}{{(x - 4)}} + \frac{{Bx + C}}{{{x^2} + 3x + 4}} \\ \Rightarrow \frac{{A({x^2} + 3x + 4) + \left( {Bx + C} \right)\left( {x - 4} \right)}}{{(x - 4)({x^2} + 3x + 4)}} = \frac{{7{x^2} - 25x + 20}}{{2(x - 4)({x^2} + 3x + 4)}} \\\end{aligned} \end{equation} $$ Equating the numerators we get, $$ \Rightarrow A({x^2} + 3x + 4) + \left( {Bx + C} \right)\left( {x - 4} \right) = \frac{{7{x^2} - 25x + 20}}{2}$$

For $(x-4)=0$ or $x=4$, $$\begin{equation} \begin{aligned} \Rightarrow 32A = 16 \\ \Rightarrow A = \frac{1}{2} \\\end{aligned} \end{equation} $$

For $x=0$, $$\begin{equation} \begin{aligned} \Rightarrow 4A - 4C = 10 \\ \Rightarrow 2 - 4C = 10\quad \quad \left( {As,\;A = \frac{1}{2}} \right) \\ \Rightarrow C = - 2 \\\end{aligned} \end{equation} $$

For $x=1$, $$\begin{equation} \begin{aligned} \Rightarrow 8A + (B + C)( - 3) = 1 \\ \Rightarrow 4 - 3B + 6 = 1\quad \quad \left( {As,\;A = \frac{1}{2}\;\ \;C = - 2} \right) \\ \Rightarrow B = 3 \\\end{aligned} \end{equation} $$

Therefore, $$ \Rightarrow \frac{{7{x^2} - 25x + 20}}{{2(x - 4)({x^2} + 3x + 4)}} = \frac{1}{{2(x - 4)}} + \frac{{3x - 2}}{{{x^2} + 3x + 4}}$$