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Your input: perform the partial fraction decomposition of $$$\frac{1}{2 x^{2} + 7 x - 4}$$$

Simplify the expression: $$$\frac{1}{2 x^{2} + 7 x - 4}=\frac{1}{\left(x + 4\right) \left(2 x - 1\right)}$$$

The form of the partial fraction decomposition is

$$\frac{1}{\left(x + 4\right) \left(2 x - 1\right)}=\frac{A}{2 x - 1}+\frac{B}{x + 4}$$

Write the right-hand side as a single fraction:

$$\frac{1}{\left(x + 4\right) \left(2 x - 1\right)}=\frac{\left(x + 4\right) A + \left(2 x - 1\right) B}{\left(x + 4\right) \left(2 x - 1\right)}$$

The denominators are equal, so we require the equality of the numerators:

$$1=\left(x + 4\right) A + \left(2 x - 1\right) B$$

Expand the right-hand side:

$$1=x A + 2 x B + 4 A - B$$

Collect up the like terms:

$$1=x \left(A + 2 B\right) + 4 A - B$$

The coefficients near the like terms should be equal, so the following system is obtained:

$$\begin{cases} A + 2 B = 0\\4 A - B = 1 \end{cases}$$

Solving it (for steps, see system of equations calculator), we get that $$$A=\frac{2}{9}$$$, $$$B=- \frac{1}{9}$$$

Therefore,

$$\frac{1}{\left(x + 4\right) \left(2 x - 1\right)}=\frac{\frac{2}{9}}{2 x - 1}+\frac{- \frac{1}{9}}{x + 4}$$

Answer: $$$\frac{1}{2 x^{2} + 7 x - 4}=\frac{\frac{2}{9}}{2 x - 1}+\frac{- \frac{1}{9}}{x + 4}$$$