Question

partial fraction decomposition -22/6x^2-7x-3

Answer 1

The fraction can be written as -22/[6*(x+1/3)(x-3/2)] where x= -1/3 and x=3/2 are the roots of 6x^2 -7x -3 = 0.

We then have: (-11/3)*[ 1/(x+1/3) * 1/(x - 3/2) ] and this we set equal to
(-11/3)*[ a/(x+1/3) + b/(x-3/2) ]. Now we have to find a and b such that the
equality holds.

We have (-11/3)*[ 1/(x+1/3) * 1/(x - 3/2) ] = (-11/3)*[ ax -3a/2 + bx + b/3]/[(x+1/3)*(x - 3/2)]
Comparing the numerators we have: 1 = x(a+b) + b/3 -3a/2. There is no x on the left side so we need: a + b = 0 or a = -b. Then we are left with:
1 = -a/3 -3a/2 or 1 = -11a/6 or a = -6/11.

The final result: -22/[6*(x+1/3)(x-3/2)] = (-11/3)*[ 1/(x+1/3) * 1/(x - 3/2) ]=
(-11/3)*[ (-6/11)* 1/(x+1/3)] + (6/11)* 1/(x-3/2) ] =
2*[ 1/(x+1/3)] - 1/(x-3/2) ].