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\[\dfrac{2x+1}{(x+1)(x-3)(x+4)} = \dfrac{A}{(x+1)}\ + \dfrac{B}{(x-3)}\ + \dfrac{C}{(x+4)}\]

The above rational fraction is expressed in the form of partial fractions. Evaluate $A + B + C$.

 Feb 9, 2021

Best Answer 

 #1
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We can use fraction decomposition, 

$A(x-3)(x+4)+B(x+1)(x+4)+C(x+1)(x-3)=2x+1$

Setting $x$ as $-1$, we get

$A(-4)(3)=-1$

$A=1/12$
 

setting $x$ as $3$, 

we get 

B(4)(7)=7

$B=1/4$
 

Setting x=-4

C(-3)(-7)=-7

$C=-1/3$
$1/12+1/4-1/3=0$

 Feb 9, 2021
 #1
avatar
+1
Best Answer

We can use fraction decomposition, 

$A(x-3)(x+4)+B(x+1)(x+4)+C(x+1)(x-3)=2x+1$

Setting $x$ as $-1$, we get

$A(-4)(3)=-1$

$A=1/12$
 

setting $x$ as $3$, 

we get 

B(4)(7)=7

$B=1/4$
 

Setting x=-4

C(-3)(-7)=-7

$C=-1/3$
$1/12+1/4-1/3=0$

Guest Feb 9, 2021

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