<PRE><B>Question 600952</B>
Write the rational expression 7/2x^2-5x-3 as a sum of partial fractions.
Factor the denominator and write it
{{{7/((2x+1)(x-3))}}} = {{{A/((2x+1))}}} + {{{B/((x-3))}}}
:
Combine the fractions on the right
{{{7/((2x+1)(x-3))}}} = {{{(A(x-3)+B(2x+1))/((2x+1)(x-3))}}}
:
If the denominators are equal, the numerators are, we can write it
7 = A(x-3) + B(2x+1)
:
Assume this is an identity, equation true for all values of x. 
Let x=3
7 = A(3-3) + B(2(3)+1)
The A term drops out
7 = 7B + 0
B = 1
:
Let x=-.5
7 = A(-.5-3) + B(2(-.5)+1)
The B term drops out
7 = -3.5A + 0
A = -2
:
The decomposed fraction
{{{7/((2x+1)(x-3))}}} = {{{(-2)/((2x+1))}}} + {{{1/((x-3))}}}


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