SOLUTION: Perform the long division and write the partial fraction decomposition of the remainder term. (x^5 + 2)/(x^2 - 1)

Algebra ->  Exponents-negative-and-fractional -> SOLUTION: Perform the long division and write the partial fraction decomposition of the remainder term. (x^5 + 2)/(x^2 - 1)       Log On


   

Question 1043286: Perform the long division and write the partial fraction decomposition of the remainder term.
(x^5 + 2)/(x^2 - 1)
Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
To do the division we insert all the necessary
zero place holders for all the missing powers
of x.  It isn't necessary to put in the 1
coefficients but I thought I'd do it anyway:

        
                1x³+0x²+1x+0
1x²+0x-1)1x⁵+0x⁴+0x³+0x²+0x+2
         1x⁵+0x⁴-1x³
             0x⁴+1x³+0x²
             0x⁴+0x³+0x²
                 1x³+0x²+0x
                 1x³+0x²-1x
                     0x²+1x+2
                     0x²+0x+0
                         1x+2


So 

%28x%5E5%2B2%29%2F%28x%5E2-1%29%22%22=%22%22x%5E3%2Bx%2B%28x%2B2%29%2F%28x%5E2-1%29

Now we break the fraction part into partial fractions:

First factor the denominator:

%28x%2B2%29%2F%28x%5E2-1%29%22%22=%22%22%28x%2B2%29%2F%28%28x-1%29%28x%2B1%29%29

Then

%28x%2B2%29%2F%28%28x-1%29%28x%2B1%29%29%22%22=%22%22A%2F%28x-1%29%2BB%2F%28x%2B1%29

Multiply through by the LCD = (x-1)(x+1)

x%2B2%22%22=%22%22A%28x%2B1%29%2BB%28x-1%29

This must be true for all values of x.

So substitute x=1 to make the last term on the right
become 0: 

1%2B2%22%22=%22%22A%281%2B1%29%2BB%281-1%29

3%22%22=%22%222A

3%2F2%22%22=%22%22A

Substitute x=-1 to make the first term on the right
become 0:

-1%2B2%22%22=%22%22A%28-1%2B1%29%2BB%28-1-1%29

1%22%22=%22%22-2B

-1%2F2%22%22=%22%22B

So 

%28x%2B2%29%2F%28%28x-1%29%28x%2B1%29%29%22%22=%22%22A%2F%28x-1%29%2BB%2F%28x%2B1%29

becomes:

%28x%2B2%29%2F%28%28x-1%29%28x%2B1%29%29%22%22=%22%22%223%2F2%22%2F%28x-1%29%2B%22-1%2F2%22%2F%28x%2B1%29

Simplify those by multiplying tops and bottoms by 2

%28x%2B2%29%2F%28%28x-1%29%28x%2B1%29%29%22%22=%22%223%2F%282%28x-1%29%29-1%2F%282%28x%2B1%29%29

Substituting in

%28x%5E5%2B2%29%2F%28x%5E2-1%29%22%22=%22%22x%5E3%2Bx%2B%28x%2B2%29%2F%28x%5E2-1%29 

%28x%5E5%2B2%29%2F%28x%5E2-1%29%22%22=%22%22x%5E3%2Bx%2B3%2F%282%28x-1%29%29-1%2F%282%28x%2B1%29%29    

Edwin