Solve-the-Partial-fraction-3x-4-9x-3-16x-2-9x-13-x-1-2-x-2-2x-2-2-

Question Number 14071 by tawa tawa last updated on 27/May/17

Solve the Partial fraction

\frac{{3 x}^{4} - {9 x}^{3} + {16 x}^{2} + 9 x + 13}{{(x - 1)}^{2} {(x^{2} + 2 x - 2)}^{2}}

Commented by Tinkutara last updated on 27/May/17

I have done by calculator :

\frac{32}{{(x - 1)}^{2}} - \frac{230}{x - 1} + \frac{93 x + 287}{{(x^{2} + 2 x - 2)}^{2}} +

\frac{230 x + 661}{x^{2} + 2 x - 2}

Answered by RasheedSindhi last updated on 28/May/17

Let

\frac{A_{1}}{x - 1} + \frac{A_{2}}{{(x - 1)}^{2}} + \frac{B_{1} x + C_{1}}{x^{2} + 2 x - 2} + \frac{B_{2} x + C_{2}}{{(x^{2} + 2 x - 2)}^{2}}

= \frac{{3 x}^{4} - {9 x}^{3} + {16 x}^{2} + 9 x + 13}{{(x - 1)}^{2} {(x^{2} + 2 x - 2)}^{2}}

Multiplying by {(x - 1)}^{2} {(x^{2} + 2 x - 2)}^{2}

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

+ (B_{1} x + C_{1}) {(x - 1)}^{2} (x^{2} + 2 x - 2)

+ (B_{2} x + C_{2}) {(x - 1)}^{2}

= {3 x}^{4} - {9 x}^{3} + {16 x}^{2} + 9 x + 13

= ({3 x}^{2} - 15 x + 52) (x^{2} + 2 x - 2) - 125 x + 117 . . (A)

Let x = 1

3 {(1)}^{4} - 9 {(1)}^{3} + 16 {(1)}^{2} + 9 (1) + 13

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

3 - 9 + 16 + 9 + 13 = A_{2} {(1 + 2 - 2)}^{2}

A_{2} = 32

Let x^{2} + 2 x - 2 = 0 \Rightarrow x = \pm \sqrt{3} - 1 = \pm a

x^{2} + 2 x - 2 = (x - a) (x + a)

putting in A

(B_{2} (\pm a) + C_{2}) {(\pm a - 1)}^{2}

= ({3 x}^{2} - 15 x + 52) (0) - 125 (\pm a) + 117

(\pm a) {(\pm a - 1)}^{2} B_{2} + {(\pm a - 1)}^{2} C_{2} = - 125 (\pm a) + 117

continue .

Commented by tawa tawa last updated on 27/May/17

Am with you sir .

Answered by RasheedSindhi last updated on 28/May/17

\frac{{3 x}^{4} - {9 x}^{3} + {16 x}^{2} + 9 x + 13}{{(x - 1)}^{2} {(x^{2} + 2 x - 2)}^{2}} \dots (A)

x^{2} + 2 x - 2 = {(x + 1)}^{2} = 3

x = \pm \sqrt{3} - 1 = \pm a

x^{2} + 2 x - 2 = (x - a) (x + a)

{3 x}^{4} - {9 x}^{3} + {16 x}^{2} + 9 x + 13

= (x^{2} + 2 x - 2) ({3 x}^{2} - 15 x + 52) - 125 x + 117

= (x - a) (x + a) ({3 x}^{2} - 15 x + 52) - 125 x + 117

(A) \Rightarrow \frac{{3 x}^{4} - {9 x}^{3} + {16 x}^{2} + 9 x + 13}{{(x - 1)}^{2} {(x - a)}^{2} {(x + a)}^{2}}

[H e r e a = \pm \sqrt{3} - 1]

\frac{{3 x}^{4} - {9 x}^{3} + {16 x}^{2} + 9 x + 13}{{(x - 1)}^{2} {(x - a)}^{2} {(x + a)}^{2}}

= \frac{A_{1}}{x - 1} + \frac{A_{2}}{{(x - 1)}^{2}} + \frac{B_{1}}{x - a} + \frac{B_{2}}{{(x - a)}^{2}} + \frac{C_{1}}{x + a} + \frac{C_{2}}{{(x + a)}^{2}}

{3 x}^{4} - {9 x}^{3} + {16 x}^{2} + 9 x + 13

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

+ A_{2} {(x - a)}^{2} {(x + a)}^{2}

+ B_{1} {(x - 1)}^{2} (x - a) {(x + a)}^{2}

+ B_{2} {(x - 1)}^{2} {(x + a)}^{2}

+ C_{1} {(x - 1)}^{2} {(x - a)}^{2} (x + a)

+ C_{2} {(x - 1)}^{2} {(x - a)}^{2}

x = 1 \Rightarrow

3 {(1)}^{4} - 9 {(1)}^{3} + 16 {(1)}^{2} + 9 (1) + 13

= A_{2} {(x - a)}^{2} {(x + a)}^{2}

3 - 9 + 16 + 9 + 13 = A_{2} {[(x - a) (x + a)]}^{2}

32 = A_{2} {(x^{2} + 2 x - 2)}^{2}

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

A_{2} = 32

x = a \Rightarrow

{3 x}^{4} - {9 x}^{3} + {16 x}^{2} + 9 x + 13

= B_{2} {(x - 1)}^{2} {(x + a)}^{2}

(x - a) (x + a) ({3 x}^{2} - 15 x + 52) - 125 x + 117

= B_{2} {(x - 1)}^{2} {(x + a)}^{2}

(0) (a + a) (3 a^{2} - 15 a + 52) - 125 a + 117

= B_{2} {(a - 1)}^{2} {(a + a)}^{2}

- 125 a + 117 = B_{2} (4 a^{2}) {(a - 1)}^{2}

B_{2} = \frac{- 125 a + 117}{4 a^{2} {(a - 1)}^{2}} = b

x = - a \Rightarrow

C_{2} {(- a - 1)}^{2} {(- a - a)}^{2}

= (x - a) (- a + a) ({3 x}^{2} - 15 x + 52) - 125 x + 117

C_{2} {(a + 1)}^{2} (4 a^{2}) = - 125 a + 117

C_{2} = \frac{- 125 a + 117}{4 a^{2} {(a + 1)}^{2}} = c

{3 x}^{4} - {9 x}^{3} + {16 x}^{2} + 9 x + 13

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

+ A_{2} {(x - a)}^{2} {(x + a)}^{2}

+ B_{1} {(x - 1)}^{2} (x - a) {(x + a)}^{2}

+ B_{2} {(x - 1)}^{2} {(x + a)}^{2}

+ C_{1} {(x - 1)}^{2} {(x - a)}^{2} (x + a)

+ C_{2} {(x - 1)}^{2} {(x - a)}^{2}

\subset ⊚ \cap ⊤ ∣ \cap \cup \in

Commented by tawa tawa last updated on 28/May/17

God bless you sir . Am with you .