How do you write the partial fraction decomposition of the rational expression $\frac{4 x - 1}{x^{2} + 3 x - 4}$ $(4x-1 )/( x^2+3x-4)$ ?

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Answer 1 · 2018-07-18T06:25:25

The answer is $= \frac{\frac{3}{5}}{x - 1} + \frac{\frac{17}{5}}{x + 4}$ $=(3/5)/(x-1)+(17/5)/(x+4)$

Perform the decomposition into partial fractions

$\frac{4 x - 1}{x^{2} + 3 x - 4} = \frac{4 x - 1}{(x - 1) (x + 4)}$ $(4x-1)/(x^2+3x-4)=(4x-1)/((x-1)(x+4))$

$= \frac{A}{x - 1} + \frac{B}{x + 4}$ $=A/(x-1)+B/(x+4)$

$= \frac{A (x + 4) + B (x - 1)}{(x - 1) (x + 4)}$ $=(A(x+4)+B(x-1))/((x-1)(x+4))$

The denominators are the same, compare the numerators

$4 x - 1 = A (x + 4) + B (x - 1)$ $4x-1=A(x+4)+B(x-1)$

Let $x = 1$ $x=1$ , $\Rightarrow$ $=>$ , $3 = 5 A$ $3=5A$ , $\Rightarrow$ $=>$ , $A = \frac{3}{5}$ $A=3/5$

Let $x = - 4$ $x=-4$ , $\Rightarrow$ $=>$ , $- 17 = - 5 B$ $-17=-5B$ , $\Rightarrow$ $=>$ , $B = \frac{17}{5}$ $B=17/5$

Therefore,

$\frac{4 x - 1}{x^{2} + 3 x - 4} = \frac{\frac{3}{5}}{x - 1} + \frac{\frac{17}{5}}{x + 4}$ $(4x-1)/(x^2+3x-4)=(3/5)/(x-1)+(17/5)/(x+4)$

How do you write the partial fraction decomposition of the rational expression 4x−1x2+3x−4 (4x-1 )/( x^2+3x-4)?