# Find the partial fractions for (2x+3)/(x^2-7x -8)

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### 1 Answer

We have to find the partial fractions for (2x + 3)/(x^2 - 7x - 8)

First factorize the denominator

x^2 - 7x - 8

=> x^2 - 8x + x - 8

=> x(x - 8) + 1(x - 8)

=> (x + 1)(x - 8)

(2x + 3)/(x^2 - 7x - 8) = A / (x + 1) + B/(x - 8)

=> Ax - 8A + Bx + B = 2x + 3

=> B - 8A = 3

and A + B = 2

=> B = 2 - A

substitute in B - 8A = 3

=> 2 - A - 8A = 3

=> -9A = 1

=> A = -1/9

B = 2 + 1/9 = 19/9

**The partial fractions are (19/9)/(x - 8) - (1/9)/(x + 1)**