Rewrite the expression f(x) = \frac {6x}{x^2-8x+15} using partial fractions.

Question:

Rewrite the expression {eq}f(x) = \frac {6x}{x^2-8x+15} {/eq} using partial fractions.

Partial Fraction Concept:

Let{eq}Q(x) {/eq} can be written as {eq}\frac{{w(x)}}{{g(x)}} {/eq} ,in which both are polynomials in x.In partial fraction degree of {eq}g(x) {/eq} must be greater than {eq}w(x) {/eq}then partial fraction for linear and quadratic the factor given by;

{eq}Q(x) = \frac{{w(x)}}{{(x + p)({x^2} + ax + b)}} {/eq}

By partial fraction expansion ,

{eq}Q(x) = \frac{A}{{(x + p)}} + \frac{{Bx + C}}{{({x^2} + ax + b)}} {/eq}

The following rules are relevant to this problem:

1.{eq}{{x^2} - 8x + 15 = \left( {x - 5} \right)\left( {x - 3} \right)} {/eq}

2.{eq}\frac{B}{B} = 1 {/eq}

3.{eq}{\frac{a}{b} + \frac{c}{d} = \frac{{ad + cb}}{{bd}}} {/eq}

Given that: {eq}\displaystyle f(x) = \frac{{6x}}{{{x^2} - 8x + 15}} {/eq}

{eq}\displaystyle \eqalign{ & f(x) = \frac{{6x}}{{{x^2} - 8x + 15}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{6x}}{{{x^2} - 5x - 3x + 15}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{6x}}{{\left( {x - 5} \right)\left( {x - 3} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{x^2} - 8x + 15 = \left( {x - 5} \right)\left( {x - 3} \right)} \right) \cr & f(x) = \frac{{6x}}{{\left( {x - 5} \right)\left( {x - 3} \right)}} \cr & {\text{Let,}} \cr & \frac{{6x}}{{\left( {x - 5} \right)\left( {x - 3} \right)}} = \frac{A}{{\left( {x - 5} \right)}} + \frac{B}{{\left( {x - 3} \right)}} \cr & \frac{{6x}}{{\left( {x - 5} \right)\left( {x - 3} \right)}} = \frac{{A\left( {x - 3} \right) + B\left( {x - 5} \right)}}{{\left( {x - 5} \right)\left( {x - 3} \right)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\frac{a}{b} + \frac{c}{d} = \frac{{ad + cb}}{{bd}}} \right) \cr & 6x = A\left( {x - 3} \right) + B\left( {x - 5} \right) \cr & \cr & {\text{Putting, }}x = 3; \cr & 6\left( 3 \right) = A\left( {3 - 3} \right) + B\left( {3 - 5} \right) \cr & 18 = A\left( 0 \right) + B\left( { - 2} \right) \cr & B = - 9 \cr & \cr & {\text{Putting, }}x = 5; \cr & 6\left( 5 \right) = A\left( {5 - 3} \right) + B\left( {5 - 5} \right) \cr & 30 = A\left( 2 \right) + B\left( 0 \right) \cr & A = 15 \cr & \cr & f(x) = \frac{{6x}}{{\left( {x - 5} \right)\left( {x - 3} \right)}} = \frac{{15}}{{\left( {x - 5} \right)}} + \frac{{\left( { - 9} \right)}}{{\left( {x - 3} \right)}} \cr & f(x) = \frac{{6x}}{{\left( {x - 5} \right)\left( {x - 3} \right)}} \cr & f(x) = \frac{{15}}{{\left( {x - 5} \right)}} - \frac{9}{{\left( {x - 3} \right)}} \cr} {/eq}