The partial fraction decomposition of $$ \frac{8 x-25}{8 x^{2}+2 x-21} $$

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We start with the expression
$$
\frac{8x-25}{8x^2+2x-21}.
$$
  
**Step 1. Factor the Denominator**

We need to factor the quadratic $$8x^2+2x-21$$. Notice that
$$
8x^2+2x-21 = (4x+7)(2x-3).
$$
You can verify this by expanding the factors:
$$
(4x+7)(2x-3)= 8x^2 -12x+14x-21 = 8x^2+2x-21.
$$

**Step 2. Set Up the Partial Fractions**

We write
$$
\frac{8x-25}{(4x+7)(2x-3)} = \frac{A}{4x+7}+\frac{B}{2x-3}.
$$

**Step 3. Clear the Denominator**

Multiply both sides of the equation by $$(4x+7)(2x-3)$$:
$$
8x-25 = A(2x-3) + B(4x+7).
$$

**Step 4. Expand and Collect Like Terms**

Expand the right-hand side:
$$
A(2x-3) + B(4x+7) = 2Ax - 3A + 4Bx + 7B = (2A+4B)x + (-3A+7B).
$$

**Step 5. Equate Coefficients**

For the equality to hold for all $$x$$, the coefficients of like powers of $$x$$ must be equal:
$$
\begin{cases}
2A+4B = 8,\[1mm]
-3A+7B = -25.
\end{cases}
$$

**Step 6. Solve the System of Equations**

From the first equation:
$$
2A+4B=8 \quad \Rightarrow \quad A = 4 - 2B.
$$
Substitute $$A = 4-2B$$ into the second equation:
$$
-3(4-2B)+7B = -25.
$$
Simplify:
$$
-12+6B+7B = -25 \quad \Rightarrow \quad -12+13B = -25.
$$
Solve for $$B$$:
$$
13B = -25+12 = -13 \quad \Rightarrow \quad B = -1.
$$
Substitute $$B = -1$$ back into $$A=4-2B$$:
$$
A =4-2(-1)=4+2=6.
$$

**Step 7. Write the Partial Fraction Decomposition**

Now substitute the values for $$A$$ and $$B$$ back into the partial fractions:
$$
\frac{8x-25}{(4x+7)(2x-3)} = \frac{6}{4x+7} - \frac{1}{2x-3}.
$$
The partial fraction decomposition is: $$ \frac{8x-25}{8x^{2}+2x-21} = \frac{6}{4x+7} - \frac{1}{2x-3}. $$

The partial fraction decomposition of \( \frac{8 x-25}{8 x^{2}+2 x-21} \)

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