(2x^2-x-19)/x(x+1)(x-1) write the partial descomposition

Question

(2x^2-x-19)/x(x+1)(x-1)

Answer 1

you need to write this as 

(2x²-x-19)/[(x+1)(x-1)] so that both linear factors belong in the denominator.

 

Since degrees of numerator & denominator are the same, perform division in order to get the problem into proper shape.

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(2x^2-x-19)/(x+1)(x-1)

Answer 2

First of all, since the degree of the numerator is equal to the degree of the denominator use long division to obtain that

 

(2x² - x - 19)/[(x+1)(x-1)] is equal to 2 + (-x - 17)/[(x+1)(x-1)].

 

Now do the decomposition for:    (-x - 17)/[(x+1)(x-1)] = A/(x+1) + B/(x - 1)

 

Multiply both sides of the equation by the common denominator (x + 1)(x - 1)

 

Thus we obtain:  (-x - 17) = A(x - 1) + B(x + 1)

 

Let x = -1:          (-(-1) - 17)) = A(-1 - 1) + B(-1 + 1)

                                      -16 = -2A

                                         8 = A

 

Let x = 1:             (-1 - 17) = A(1 - 1) + B(1 + 1)

                                       -18 = 2B

                                          -9 = B

 

Thus,  (2x² - x - 19) = 2 + 8/(x + 1) - 9/(x - 1)