Precalculus (6th Edition) Blitzer

The simplified partial fraction expansion is $\frac{6{{x}^{2}}-14x+27}{\left( x+2 \right){{\left( x-3 \right)}^{2}}}=\frac{A}{\left( x+2 \right)}+\frac{B}{\left( x-3 \right)}+\frac{C}{{{\left( x-3 \right)}^{2}}}$.
The provided rational expression is as follows: $\frac{6{{x}^{2}}-14x+27}{\left( x+2 \right){{\left( x-3 \right)}^{2}}}$ Now, solving the expression as follows: We set up the partial fraction expansion with unknown constants coefficients and then write a constant coefficients over each of the two distinct algebraic linear factors in the denominator of the expression. $\frac{6{{x}^{2}}-14x+27}{\left( x+2 \right){{\left( x-3 \right)}^{2}}}=\frac{A}{\left( x+2 \right)}+\frac{B}{\left( x-3 \right)}+\frac{C}{{{\left( x-3 \right)}^{2}}}$ Thus, we get $\frac{A}{\left( x+2 \right)}+\frac{B}{\left( x-3 \right)}+\frac{C}{{{\left( x-3 \right)}^{2}}}$ is a partial fraction expansion of the rational expression $\frac{6{{x}^{2}}-14x+27}{\left( x+2 \right){{\left( x-3 \right)}^{2}}}$ with constants $A$, $B$ and $C$.