## Precalculus (6th Edition) Blitzer

The partial fraction is, $\frac{3}{\left( x-4 \right)}+\frac{2x-1}{\left( {{x}^{2}}+5 \right)}$. $\frac{5{{x}^{2}}-9x+19}{\left( x-4 \right)\left( {{x}^{2}}+5 \right)}=\frac{A}{\left( x-4 \right)}+\frac{Bx+C}{\left( {{x}^{2}}+5 \right)}$
Now, take L.C.M on the right side: $\frac{5{{x}^{2}}-9x+19}{\left( x-4 \right)\left( {{x}^{2}}+5 \right)}=\frac{A\left( {{x}^{2}}+5 \right)+\left( Bx+C \right)\left( x-4 \right)}{\left( x-4 \right)\left( {{x}^{2}}+5 \right)}$. By eliminating the denominators from both sides, we get: \begin{align} & 5{{x}^{2}}-9x+19=A\left( {{x}^{2}}+5 \right)+\left( Bx+C \right)\left( x-4 \right) \\ & =A{{x}^{2}}+5A+B{{x}^{2}}-4Bx+Cx-4C \\ & =\left( A+B \right){{x}^{2}}+\left( C-4B \right)x+5A-4C \end{align} Then, compare the coefficient of ${{x}^{2}},\ x$ and the constant term $A+B=5$ …… (I) $C-4B=-9$ …… (II) $5A-4C=19$ …… (III) Equation (I): multiply by 4 and add with equation (II): \begin{align} & 4A+4B=20 \\ & C-4B=-9 \end{align} $4A+C=11$ …… (IV) Equation (IV) multiplies by 4 and adds with equation (III): \begin{align} & 16A+4C=44 \\ & 5A-4C=19 \\ & 21A=\ 63 \\ & A=3 \end{align} Then, put the value of A in equation (IV): \begin{align} & 12+C=11 \\ & C=-1 \end{align} Also, put the value of A in equation (I) \begin{align} & 3+B=5 \\ & B=2 \end{align} Then, $\frac{5{{x}^{2}}-9x+19}{\left( x-4 \right)\left( {{x}^{2}}+5 \right)}=\frac{3}{\left( x-4 \right)}+\frac{2x-1}{\left( {{x}^{2}}+5 \right)}$ Thus, the partial fraction of the provided expression is $\frac{3}{\left( x-4 \right)}+\frac{2x-1}{\left( {{x}^{2}}+5 \right)}$.