Decompose the following rational expression into partial fractions: (2x-7)/(25x - 24 - 6x2)

The denominator is $25x - 24 - 6x^2$. Rearrange it into standard quadratic form:

$$-6x^2 + 25x - 24$$

Factor out $-1$:

$$-(6x^2 - 25x + 24)$$

Now, factorize the quadratic expression $6x^2 - 25x + 24$. We look for two numbers that multiply to $6 \times 24 = 144$ and add up to $-25$. These numbers are $-9$ and $-16$.

$$6x^2 - 9x - 16x + 24$$

Factor by grouping:

$$3x(2x - 3) - 8(2x - 3)$$ $$(3x - 8)(2x - 3)$$

So, the denominator is $-(3x - 8)(2x - 3)$, which can also be written as $(8 - 3x)(2x - 3)$.
The expression becomes:

$$\frac{2x-7}{(8-3x)(2x-3)}$$

We express the fraction as a sum of two simpler fractions:

$$\frac{2x-7}{(8-3x)(2x-3)} = \frac{A}{8-3x} + \frac{B}{2x-3}$$

Multiply both sides by $(8-3x)(2x-3)$ to clear the denominators:

$$2x-7 = A(2x-3) + B(8-3x)$$

To find A, set $2x-3 = 0$, which means $x = \frac{3}{2}$:

$$2\left(\frac{3}{2}\right) - 7 = A\left(2\left(\frac{3}{2}\right) - 3\right) + B\left(8 - 3\left(\frac{3}{2}\right)\right)$$ $$3 - 7 = A(3 - 3) + B\left(8 - \frac{9}{2}\right)$$ $$-4 = A(0) + B\left(\frac{16}{2} - \frac{9}{2}\right)$$ $$-4 = B\left(\frac{7}{2}\right)$$ $$B = -4 \times \frac{2}{7} = -\frac{8}{7}$$

To find B, set $8-3x = 0$, which means $3x = 8$, so $x = \frac{8}{3}$:

$$2\left(\frac{8}{3}\right) - 7 = A\left(2\left(\frac{8}{3}\right) - 3\right) + B\left(8 - 3\left(\frac{8}{3}\right)\right)$$ $$\frac{16}{3} - 7 = A\left(\frac{16}{3} - \frac{9}{3}\right) + B(8 - 8)$$ $$\frac{16}{3} - \frac{21}{3} = A\left(\frac{7}{3}\right) + B(0)$$ $$-\frac{5}{3} = A\left(\frac{7}{3}\right)$$ $$A = -\frac{5}{3} \times \frac{3}{7} = -\frac{5}{7}$$

Substitute the values of A and B back into the partial fraction form:

$$\frac{2x-7}{25x-24-6x^2} = \frac{-\frac{5}{7}}{8-3x} + \frac{-\frac{8}{7}}{2x-3}$$

This can be written as:

$$-\frac{5}{7(8-3x)} - \frac{8}{7(2x-3)}$$