How do you write the partial fraction decomposition of the rational expression $(4x+23)/(x^2-x-6)$ ?

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Answer 1 · 2015-12-25T04:05:16

$(4x+23)/(x^2-x-6)=7/(x-3)-3/(x+2)$

Factor the denominator and split apart the expression.

$(4x+23)/((x-3)(x+2))=A/(x-3)+B/(x+2)$

Find a common denominator of $(x-3)(x+2)$ .

$4x+23=A(x+2)+B(x-3)$

Plug in $-2$ for $x$ :

$4(-2)+23=B(-2-3)$
$B=-3$

Plug in $3$ for $x$ :

$4(3)+23=A(3+2)$
$A=7$

Plug this into the original expression:

$(4x+23)/(x^2-x-6)=7/(x-3)-3/(x+2)$

How do you write the partial fraction decomposition of the rational expression (4x+23)/(x^2-x-6) (4x+23)/(x^2-x-6)?