Answer:
[tex]\dfrac{-11}{4(x+3)}+\dfrac{1}{12(x-1)}+\dfrac{8}{3(x+2)}[/tex]
Step-by-step explanation:
[tex]\dfrac{3x - 2}{(x + 3)(x - 1)(x + 2)}=\dfrac{A}{x+3}+\dfrac{B}{x-1}+\dfrac{C}{x+2}\\\\A=\dfrac{3(-3)-2}{(-3-1)(-3+2)}=\dfrac{-11}{4}\\\\B=\dfrac{3(1)-2}{(1+3)(1+2)}=\dfrac{1}{12}\\\\C=\dfrac{3(-2)-2}{(-2+3)(-2-1)}=\dfrac{-8}{-3}=\dfrac{8}{3}\\\\\dfrac{3x - 2}{(x + 3)(x - 1)(x + 2)}=\boxed{\dfrac{-11}{4(x+3)}+\dfrac{1}{12(x-1)}+\dfrac{8}{3(x+2)}}[/tex]
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When the denominator factors are linear and to the first degree, the corresponding coefficient can be found by eliminating that factor and evaluating the expression for the value of x that would make the factor zero.
As we showed above, the coefficient A is found by evaluating the expression for x=-3 (the zero of the denominator of A) with that factor eliminated from the denominator. Likewise for the others.
