Question 1162547
<pre>
Write the partial fraction decomposition of the following rational expression.
{{{ (8x^2+10x+20)/x(x+4)(x+5) }}}

The other person is WRONG!!

{{{(8x^2 + 10x + 20)/x(x + 4)(x + 5)}}} = {{{A/x + B/(x + 4) + C/(x + 5)}}}
{{{(8x^2 + 10x + 20)/x(x + 4)(x + 5)}}} = {{{A(x + 4)(x + 5)/x + B(x)(x + 5)/(x + 4) + C(x)(x + 4)/(x + 5)}}} --- Multiplying right-side by LCD, x(x + 4)(x + 5)

{{{8x^2 + 10x + 20 = A(x + 4)(x + 5) + B(x)(x + 5) + C(x)(x + 4)}}} --- Equating NUMERATORS, since DENOMINATORS are same
{{{8(- 5)^2 + 10(- 5) + 20 = A(x + 4)(- 5 + 5) + B(x)(- 5 + 5) + C(- 5)(- 5 + 4)}}} ------ Substituting - 5 for x, to determine the value of C
8(25) - 50 + 20 = 0 + 0 + C(- 5)(- 5 + 4)
  200 - 50 + 20 = C(- 5)(- 1)
            170 = 5C
           {{{170/5}}} = 34 = C

{{{8x^2 + 10x + 20 = A(x + 4)(x + 5) + B(x)(x + 5) + C(x)(x + 4)}}}
{{{8(- 4)^2 + 10(- 4) + 20 = A(- 4 + 4)(x + 5) + B(- 4)(- 4 + 5) + C(x)(- 4 + 4)}}} ------ Substituting - 4 for x, to determine the value of B
8(16) - 40 + 20 = 0 + B(- 4)(- 4 + 5) + 0
  128 - 40 + 20 = B(- 4)(1)
            108 = - 4B
          {{{108/(- 4)}}} = - 27 = B

{{{8x^2 + 10x + 20 = A(x + 4)(x + 5) + B(x)(x + 5) + C(x)(x + 4)}}}
{{{8(1)^2 + 10(1) + 20 = A(1 + 4)(1 + 5) + (- 27)(1)(1 + 5) + 34(1)(1 + 4)}}} ------ Substituting 1 for x, 34 for C, and - 27 for B, to determine the value of A
8(1) + 10 + 20 = A(5)(6) - 27(1)(6) + 34(1)(5)
   8 + 10 + 20 = 30A - 162 + 170
            38 = 30A + 8
        38 - 8 = 30A
            30 = 30A
           {{{30/30)}}} = 1 = A

So, (A, B, C)  = (1, - 27, 34)

We then get: {{{highlight((8x^2 + 10x + 20)/x(x + 4)(x + 5))}}} = {{{A/x + B/(x + 4) + C/(x + 5)}}} = {{{highlight(1/x + (- 27)/(x + 4) + 34/(x + 5))}}}</pre>