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Resolve `(3x+5)/((x+2)(3x-1))` into partial fractions. |
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Answer» In the given fraction, the denominator has two linear,non-repeated factors. `:.` The given fraction can be written as the sum of two partial fractions. Let `(3x+5)/((x+2)(3x-1))=(A)/(x+2)+(B)/(3x-1)` `implies3x+5=A(3x-1)+B(x+2)` .......`(1)` Put `x=-2` in Eq. `(1)`, `implies3(-2)+5=A[3(-2)-1]+B(-2+2)-1=-7AimpliesA=(1)/(7)` Comparing the costant the constant terms on both sides of Eq. `(1)`, we have `5=-A+2B` `5=-(1)/(7)+2B` or `B=(18)/(7)` `:. (3x+5)/((x+2)(3x-1))=((1)/(7))/(x+2)+((18)/(7))/(3x-1)=(1)/(7)[(1)/(x+2)+(18)/(3x-1)]` |
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