express-2x-2-x-2-x-2-2-1-2x-as-partial-fraction-

Question Number 14421 by chux last updated on 31/May/17

express \frac{{2 x}^{2} - x + 2}{{(x + 2)}^{2} (1 - 2 x)} as partial

fraction

Answered by RasheedSindhi last updated on 01/Jun/17

Let

\frac{{2 x}^{2} - x + 2}{{(x + 2)}^{2} (1 - 2 x)} = \frac{A_{1}}{x + 2} + \frac{A_{2}}{{(x + 2)}^{2}} + \frac{B}{1 - 2 x}

{2 x}^{2} - x + 2 = A_{1} (x + 2) (1 - 2 x) + A_{2} (1 - 2 x) + B {(x + 2)}^{2}

This is an identity i - e {it}^{'} s equal

for all values of x .

Let x = - 2

2 {(- 2)}^{2} - (- 2) + 2 = A_{2} (1 - 2 (- 2))

A_{2} = \frac{12}{5}

Let x = 1 / 2

2 {(1 / 2)}^{2} - (1 / 2) + 2 = B {(\frac{1}{2} + 2)}^{2}

B (\frac{25}{4}) = \frac{1}{2} - \frac{1}{2} + 2

B = 2 (\frac{4}{25}) = \frac{8}{25}

To determine A_{1} :

{2 x}^{2} - x + 2 = A_{1} (x + 2) (1 - 2 x) + A_{2} (1 - 2 x) + B {(x + 2)}^{2}

Substitute A_{2} = \frac{12}{5} & B = \frac{8}{25} in above .

{2 x}^{2} - x + 2 = A_{1} (x + 2) (1 - 2 x) + (\frac{12}{5}) (1 - 2 x) + (\frac{8}{25}) {(x + 2)}^{2}

{25 x}^{2} - 25 x + 50 = {25 A}_{1} (- {2 x}^{2} - 3 x + 2)

+ 60 (1 + 2 x) + 8 (x^{2} + 4 x + 4)

{25 x}^{2} - 25 x + 50 = - {50 A}_{1} x^{2} - {75 A}_{1} x + {50 A}_{1} + 60 + 120 x + {8 x}^{2} + 32 x + 32

{25 x}^{2} - 25 x + 50 = (- {50 A}_{1} + 8) x^{2} - {75 A}_{1} x + {50 A}_{1} + 60 + 120 x + 32 x + 32

\subset \oplus \cap ⊤! \cap \cup \in \dots

Commented by Tinkutara last updated on 31/May/17

Sorry, but A_{1} = \frac{- 21}{25}, A_{2} = \frac{12}{5} and

B = \frac{8}{25} (By calculator)

Commented by chux last updated on 01/Jun/17

thank you very much sir

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