How do you write the partial fraction decomposition of the rational expression $\frac{x + 10}{x^{2} + 2 x - 8}$ $(x+10)/(x^2+2x-8)$ ?

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Answer 1 · 2015-12-09T07:35:48

The equivalent partial fraction is :
$\frac{- 1}{x + 4} + \frac{2}{x - 2}$ $(-1)/(x+4) + 2/(x-2)$

Given $\frac{x + 10}{x^{2} + 2 x - 8}$ $(x+10)/(x^2 +2x-8)$

Step 1: Factor the denominator

$\frac{x + 10}{(x + 4) (x - 2)}$ $(x+10)/((x+4)(x-2)$

Step 2: Set up the partial faction as follows:

$\frac{x + 10}{(x + 4) (x - 2)} = \frac{A}{x + 4} + \frac{B}{x - 2} (1)$ $(x+10)/((x+4)(x-2)) = A/(x+4) + B/(x-2) " " " " " (1)$

Step 3: Multiply both sides by the LCD, $(x + 4) (x - 2)$ $(x+4)(x-2)$ :

$(x + 10) = A (x - 2) + B (x + 4)$ $(x+10) = A(x-2) +B(x+4)$
$x + 10 = A x - 2 A + B x + 4 B$ $x+ 10 = Ax - 2A + Bx+ 4B$

Step 4: Set up a system like this
$1 x : A + B = 1 (2)$ $1x: " " " " A+ B= 1 " " " "(2)$
$10 : -2A+4B= 10 (3)$ $10: " " " -2A+4B= 10 " " " "(3)$

Step 5. You can solve the system by the elimination method:

$2 (A + B = 1) \Rightarrow 2 A + 2 B = 2$ $2(A+B= 1) => 2A + 2B= 2$

$+ - 2 A + 4 B = 10$ $+ -2A + 4B= 10$
$6 B = 12 \Rightarrow B = 2$ $6B = 12 => B= 2$

Solve for $A$ $A$ by substituting $B = 3$ $B = 3$ into $(2)$ $(2)$ :

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

Step 6. Substitute $A$ $A$ and $B$ $B$ back into $(1)$ $(1)$ :

$\frac{x + 10}{(x + 4) (x - 2)} = \frac{- 1}{x + 4} + \frac{2}{x - 2}$ $(x+10)/((x+4)(x-2))= (-1)/(x+4) + 2/(x-2)$

How do you write the partial fraction decomposition of the rational expression (x+10)/(x^2+2x-8)x+10x2+2x−8 (x+10)/(x^2+2x-8)?