+0

+1
49
3

Find constants A, B, C so that Enter your answer in the form "A, B, C".

Apr 25, 2020

#3
+1

[ 2x2 + 2x - 2 ] / [ x(x2 - 1) ]  =  [ 2x2 + 2x - 2 ] / [ x(x + 1)(x - 1) ]

To get each term to have the common denominator of x(x + 1)(x - 1),

--  multiply the first term:  A/x  by  [ (x + 1)(x - 1) ] / [ (x + 1)(x - 1)]

so that the numerator becomes:  A(x + 1)(x - 1)  =  Ax2 - A

--  multiply the second term:  B/(x - 1)  by  [ x(x + 1) ] / [ x(x + 1) ]

so that the numerator becomes:  B(x)(x + 1)  =  Bx2 + Bx

--  multiply the third term:  C/(x + 1)  by  [ x(x - 1) ] / [ x(x - 1) ]

so that the numerator becomes:  C(x)(x - 1)  =  Cx2 - Cx

All the denominator are  (x)(x + 1)(x - 1), so all we have to do is look at the numerators:

2x2 + 2x - 2  =   Ax2 - A  +  Bx2 + Bx  +  Cx2 - Cx

2x2 + 2x - 2  =   Ax2 +  Bx2 +  Cx2  + BX - Cx - A

Therefore:  Ax2 +  Bx2 +  Cx2  =  2x2      --->     (A + B + C)x2  =  2x2     --->     A + B + C  =  2

Bx - Cx  =  2x       --->               (B - C)x  =  2x      --->            B - C  =  2

-A  =  -2                                                        --->                  A  =  2

Solving, we get:  A = 2     B = 1     C = -1

Apr 25, 2020