How do you express $\frac{x^{2} - 3 x + 2}{4 x^{3} + 11 x^{2}}$ $(x^2 - 3x + 2) / (4x^3 + 11x^2)$ in partial fractions?

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How do you express $\frac{x^{2} - 3 x + 2}{4 x^{3} + 11 x^{2}}$ $(x^2 - 3x + 2) / (4x^3 + 11x^2)$ in partial fractions?

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Answer 1 · 2016-12-26T13:27:12

The answer is $= \frac{\frac{2}{11}}{x^{2}} + \frac{- \frac{41}{121}}{x} + \frac{\frac{285}{121}}{4 x + 11}$ $=(2/11)/x^2+(-41/121)/x+(285/121)/(4x+11)$

Explanation:

We start the decomposition into partial fractions

$\frac{x^{2} - 3 x + 2}{4 x^{3} + 11 x^{2}} = \frac{x^{2} - 3 x + 2}{x^{2} (4 x + 11)}$ $(x^2-3x+2)/(4x^3+11x^2)=(x^2-3x+2)/(x^2(4x+11))$

$= \frac{A}{x^{2}} + \frac{B}{x} + \frac{C}{4 x + 11}$ $=A/x^2+B/x+C/(4x+11)$

$= \frac{A (4 x + 11) + B x (4 x + 11) + C x^{2}}{x^{2} (4 x + 11)}$ $=(A(4x+11)+Bx(4x+11)+Cx^2)/(x^2(4x+11))$

Therefore,

$x^{2} - 3 x + 2 = A (4 x + 11) + B x (4 x + 11) + C x^{2}$ $x^2-3x+2=A(4x+11)+Bx(4x+11)+Cx^2$

Let, $x = 0$ $x=0$ , $\Rightarrow$ $=>$ , $2 = 11 A$ $2=11A$ , $\Rightarrow$ $=>$ , $A = \frac{2}{11}$ $A=2/11$

Coefficients of $- 3 = 4 A + 11 B$ $-3=4A+11B$

$11 B = - 3 - 4 A = - 3 - \frac{8}{11}$ $11B=-3-4A=-3-8/11$

$B = - \frac{41}{121}$ $B=-41/121$

Coefficients of $x^{2}$ $x^2$ , $1 = 4 B + C$ $1=4B+C$

$C = 1 - 4 B$ $C=1-4B$

$C = 1 + \frac{164}{121}$ $C=1+164/121$

$C = \frac{285}{121}$ $C=285/121$

So,

$\frac{x^{2} - 3 x + 2}{4 x^{3} + 11 x^{2}} = \frac{\frac{2}{11}}{x^{2}} + \frac{- \frac{41}{121}}{x} + \frac{\frac{285}{121}}{4 x + 11}$ $(x^2-3x+2)/(4x^3+11x^2)=(2/11)/x^2+(-41/121)/x+(285/121)/(4x+11)$

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How do you express $\frac{x^{2} - 3 x + 2}{4 x^{3} + 11 x^{2}}$ $(x^2 - 3x + 2) / (4x^3 + 11x^2)$ in partial fractions?

1 Answer

Explanation:

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