Question

partial fraction
what are the steps for getting A B C values out their?​

Answer 1

Step-by-step explanation:

We have,

$\rm \frac{ {x}^{2} + 1}{ {x}^{3} + 3 {x}^{2} + 3x + 2 } = \frac{ A }{x + 2} + \frac{B}{ {x}^{2} + x + 1 } + \frac{C}{(x + 1)( {x}^{2} + x + 1) }$

$\rm \implies \frac{ {x}^{2} + 1}{ {x}^{3} + 3 {x}^{2} + 3x + 2 } = \frac{ A( {x}^{2} + x + 1) + B(x + 2) + C}{(x + 2)( {x}^{2} + x + 1) }$

$\rm \implies \frac{ {x}^{2} + 1}{ {x}^{3} + 3 {x}^{2} + 3x + 2 } = \frac{ A {x}^{2} +Ax + A + Bx + 2B + C}{ {x}^{2} (x + 2) + x (x + 2)+ 1(x + 2) }$

$\rm \implies \frac{ {x}^{2} + 1}{ {x}^{3} + 3 {x}^{2} + 3x + 2 } = \frac{ A {x}^{2} +Ax + Bx +A+ 2B + C}{ {x}^{3} + 2 {x}^{2} + {x}^{2} + 2x+ x + 2 }$

$\rm \implies \frac{ {x}^{2} + 1}{ {x}^{3} + 3 {x}^{2} + 3x + 2 } = \frac{ A {x}^{2} +(A + B)x +(A+ 2B + C)}{ {x}^{3} + 3{x}^{2} + 3x+ 2 }$

$\rm \implies {x}^{2} + 1 = A {x}^{2} +(A + B)x +(A+ 2B + C)$

On comparing the cooeficients,

$A = 1 \\ A + B = 0 \\ A + 2B + C = 1$

$\implies 1 + B = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \:\\ \implies \ 1 + 2B + C = 1$

$\implies B = - 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \:\\ \implies \ 1 + 2B + C = 1$

So,

$1+ 2( - 1)+ C = 1$

$\implies - 2+ C = 0$

$\implies C = 2$

So,

$\tt \purple{A - B + C = (1) - ( - 1) + 2}$

$\tt \purple{ \implies \: A - B + C = 1 + 1 + 2 = 4}$