Find out the partial fraction decomposition of each of the following.
8x^{2} -12/( x( x^{2} + 2 x - 6)
Solution
In this case the x which sits in the front is a linear term as we can write it as, x= x + 0 and thus the form of the partial fraction decomposition is,
8x^{2} -12 /x( x^{2} + 2 x - 6) = A /x + Bx+ C/ x^{2} + 2 x - 6
Now we'll utilizes the fact that the LCD is and add the two terms together,
8x^{2} -12 / x( x^{2} + 2 x - 6) = A ( x^{2} + 2 x - 6 )+ x ( Bx+ C ) / x(x^{2} + 2 x - 6)
After that, set the numerators equal.
8x^{2 }-12 = A ( x^{2} + 2 x - 6 )+ x ( Bx+ C )
This is where the procedure changes from the previous example. We could select x= 0 to obtain the value of A, although that's the only constant which we could get by using this method and thus it just won't work all that well here.
What we have to do here is multiply the right side out & then gather all the like terms as follows,
8x^{2} -12 = Ax^{2} + 2 Ax - 6 A + Bx^{2} + Cx
8x^{2} -12 = ( A + B )x^{2 }+ ( 2 A + C ) x - 6 A
Now, we have to choose A, B, and C hence that these two are equal. This means that the coefficient of the x^{2} term on the right side will need to be 8 as that is the coefficient of the x^{2} term on the left side. Similarly, the coefficient of the x term on the right side has to be zero as there isn't an x term on the left side. At last the constant term on the right side has to be -12 as that is the constant on the left side.
Generally we call these setting coefficients equivalent & we'll write down the following equations.
A + B= 8
2 A + C = 0
-6 A= -12
Now, we haven't talked regarding how to solve systems of equations still, although this is one that we can do without that knowledge. We can solve out the third equation directly for A to obtain that A = 2 . Then we can plug this into the first two equations to obtain,
2 + B = 8 ⇒ B = 6
2 ( 2) + C = 0 ⇒ C = -4
Thus, the partial fraction decomposition for this expression is,
8x^{2} -12/ x( x^{2} + 2 x - 6) = (2/x)+(6x-4)/ ( x^{2} + 2 x - 6)