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Resolve `(2x^(2)-5x+7)/((x+1)^(2)(x+3)(2x+1))` into partial fractions. |
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Answer» Let `(2x^(2)-5x+7)/((x+1)^(2)(x+3)(2x+1))=(A)/(x+1)+(B)/((x+1)^(2))+(C )/(x+3)+(D)/(2x+1)` `implies2x^(2)-5x+7=A(x+1)(x+3)(2x+1)+B(x+3)(2x+1)+C(x+1)^(2)(2x+1)+D(x+1)^(2)(x+3)`…………..`(1)` Substituting `x=-1` in Eq. `(1)`, we have `2(-1)^(2)-5(-1)+7=A(0)+B(-1+3)(-2+1)+C(0)+D(0)14=-2B` `B=-7` Substituting `x=-3` in Eq. `(1)` , we get `40=A(0)+B(0)+C(-20)+D(0)-20C=40` `C=-2` Substituting `x=-(1)/(2)` in Eq. `(1)` , we have `10=(5)/(8)DimpliesD=16` Substituting `x=0`, we have `7=3A+3B+C+3D` Substituting the values of `B,C,D` in the above equation , we get `A=-6` `:.(2x^(2)-5x+7)/((x+1)^(2)(x+3)(2x+1))=(-6)/(x+1)+(-7)/((x+1)^(2))+(-2)/(x+3)+(16)/(2x+1)` |
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