Question 1146014
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1. (a) resolve into partial fractions 2x/[(x-2)(x+5)]
   (b) resolve into partial fractions 1/(x^2-x)

<font color = red><font size = 4><b>1. (a)</font></font></b> resolve into partial fractions {{{2x/(x - 2)(x + 5)}}}
The other person’s solution: A = {{{2/3}}}, and B = {{{4/3}}}, is WRONG!!

{{{2x/(x - 2)(x + 5)}}} = {{{A/(x - 2) + B/(x + 5)}}}
{{{matrix(1,3, 2x/(x  -  2)(x + 5), "=", (A(x + 5) + B(x - 2))/(x - 2)(x + 5))}}} ---- Multiplying by LCD, (x - 2)(x + 5)
           2x = A(x + 5) + B(x - 2) ---- Equating NUMERATORS, since denominators are the same
         2(2) = A(2 + 5) + B(2 - 2) ---- Substituting 2 for x to determine the value of A
            4 = 7A
                {{{4/7 = A}}}
       2(- 5) = A(- 5 + 5) + B(- 5 - 2) ---- Substituting - 5 for x to determine the value of B
         - 10 = - 7B
        {{{(- 10)/(- 7) = 10/7 = B}}}
       (A, B) = ({{{4/7}}}, {{{10/7}}})

Therefore, {{{highlight(2x/(x - 2)(x + 5))}}} = {{{A/(x - 2) + B/(x + 5)}}} = {{{highlight((4/7)/(x - 2) + (10/7)/(x + 5))}}}


<font color = red><font size = 4><b>1. (b)</font></font></b> resolve into partial fractions {{{1/(x^2 - x)}}}

Use the same concept above, to decompose this PROPER FRACTION too. 

Before doing so though, we FACTORIZE the denominator in {{{1/(x^2 - x)}}} to get: {{{1/x(x - 1)}}}, and then: {{{1/x(x - 1)}}} = {{{A/x + B/(x - 1)}}}</pre>