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Find the partial fractions of (x^2+x-3)/((x-2)(x^2+2))
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Using partial fractions we can write the term as;
`(x^2+x-3)/((x-2)(x^2+2)) = A/(x-2)+(Bx+C)/(x^2+2)`
`(x^2+x-3)/((x-2)(x^2+2)) = (A(x^2+2)+(Bx+C)(x-2))/((x-2)(x^2+2))`
`x^2+x-3 = Ax^2+2A+(Bx+C)(x-2)`
`x^2+x-3 = Ax^2+2A+Bx^2-2Bx+Cx-2C`
`x^2+x-3 = (A+B)x^2+(C-2B)x+2A-2C`
`x^2 rarr 1 = A+B ---(1)`
`x rarr 1 = C-2B ---(2)`
cons. `rarr -3 = 2A-2C ----(3)`
`2+1 = 2A+C`
`3 = 2A+C ----(4)`
Solving (3) and (4) gives;
`C = 2`
`A = 1/2`
From (1) we can get `B = 1/2`
So the answer is;
`(x^2+x-3)/((x-2)(x^2+2)) = (1/2)/(x-2)+((1/2)x+2)/(x^2+2)`
Posted by jeew-m on May 8, 2013 at 5:57 AM (Answer #1)
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