Question

Resolve `(1)/((x-4)(x^(2)+3))` into partical fractions.
A. `(1)/(19)[(1)/(x-4)+(x+4)/(x^(2)+3)]`
B. `(1)/(19)[(1)/(x-4)+(x+3)/(x^(2)+3)]`
C. `(1)/(19)[(1)/(x-4)+(x+3)/(x^(2)+3)]`
D. `(1)/(19)[(1)/(x-4)-(x+4)/(x^(2)+3)]`

Answer 1

`(1)/((x-4)(x^(2)+3))=(A)/(x-4)+(B)/(x^(2)+3)`
`(1)/((x-4)(x^(2)+3))`
`=(A(x^(2)+3)+(x-4)(Bx+C))/((x-4)(x^(2)+3))`
Consider, `Ax^(2)+3A+bx^(2)+cx-4Bx-4C=1`
`(A+B)x^(2)+(C-4B)x+3A-4C=1`
Comparing the like terms, we get
`A+B=0`, `C-4B=0` and `(3A-4C)=1`
Solving the above equations, we get
`A=(1)/(19)`, `B=(-1)/(19)`, `C=(-4)/(19)`.
`:.(1)/((x-4)(x^(2)+3))=(1)/(19)[(1)/(x-4)-(x+4)/(x^(2)+3)]`