^{1}/

_{x+3}-

^{x}/

_{x-2}+

^{x^2+2}/

_{x^2-x-2 }(used super/subscript and "/" to show fraction. Not sure if that's proper format)

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Jose R. | Proficient and Patient Math TutorProficient and Patient Math Tutor

first factor x^{2}-x-2 to get (x-2)(x+1). The reason I factored it was to determine what's going to be my LCD. With that being said my LCD therefore is: LCD: (x+3)(x-2)(x+1). Notice that (x-2) was a repeated factor but you only account for it once.

...1/x+3 goes into the LCD (x+3) times so you multiply the numerator which is 1 by the remaining factors (x-2)(x+1). we get (x+2)(x+1)/LCD...x^{2}-x-2/LCD

...x/x-2 goes in the LCD (x-2) ties so u multiply the numerator x by (x+3)(x+1). we get x(x+3)(x+1). multiply everything out to get x^{3}-4x^{2}-3x/LCD.(remember to multiply by negative 1)

...x^{2}+2/x^{2}-x-2 goes in the LCD (x-2)(x+1) times so multiply the numerator x^{2}+2 times the remaining factor x+3.

we get. x^{3}+3x^{2}+2x+6.

Now that we have found the LCD we can perform the indicated operations. so the final answer after collecting like terms is...-2x+4/LCD.. the cubic and quadratic factors will cancel out.

now in your numerator, divide it by -2 to get x-2.

so we no have x-2/(x+3)(x+1)(x-2)

Notice that (x-2) cancels so the final answer is ......................1/(x+3)(x+1)

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