We have to write (2x-5)/(x^2-7x-8) as partial fractions.

First find the factors of the denominator.

x^2 - 7x - 8

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

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

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

(2x-5)/(x^2-7x-8) = A / (x + 1) + B / (x - 8)

= 2x - 5 = A(x - 8) + B(x + 1)

=> 2x - 5 = Ax - 8A + Bx + B

=> A + B = 2 and -8A + B = -5

subtract the two equations

=> 9A = 7

=> A = 7/9

B = 2 - 7/9 = 11/9

**The partial fractions are 7/9*(x + 1) + 11/9*(x - 8)**

We need to rewrite into partial fractions.

First we will factor the denominator.

==> x^2 - 7x -8 = (x-8)(x+1)

==> (2x-5)/x^2 -7x -8 = A/(x-8) + B(x+1)

Now we will multiply by (x-8)(x+1)

==> 2x -5 = A(x+1) + B(x-8)

==> 2x -5 = Ax + A + Bx - 8B

==> 2x-5 = (A+B)x + (A-8B)

==> A+B = 2 ==> A = 2-B

A-8B = -5

==> (2-B) - 8B = -5

==> -9B = -7

==> B = 7/9

==> A = 2- 7/9 = 11/9

**==> (2x-5)/x^2-7x-8 = 11/9(x-8) + 7/9(x+1)**