$2x+3+\frac{ 8x-4 }{ 3{x}^{2}-2x-1 }$
Write $-2x$ as a difference$2x+3+\frac{ 8x-4 }{ 3{x}^{2}+x-3x-1 }$
Factor out $x$ from the expression$2x+3+\frac{ 8x-4 }{ x \times \left( 3x+1 \right)-3x-1 }$
Factor out the negative sign from the expression$2x+3+\frac{ 8x-4 }{ x \times \left( 3x+1 \right)-\left( 3x+1 \right) }$
Factor out $3x+1$ from the expression$2x+3+\frac{ 8x-4 }{ \left( 3x+1 \right) \times \left( x-1 \right) }$
Prepare for partial-fraction decomposition by isolating the fraction$\frac{ 8x-4 }{ \left( 3x+1 \right) \times \left( x-1 \right) }$
For each factor in the denominator, write a new fraction using the factors as new denominators. The numerators are unknown values$\frac{ ? }{ 3x+1 }+\frac{ ? }{ x-1 }$
Since the factor in the denominator is linear, the numerator is an unknown constant $A$$\frac{ A }{ 3x+1 }+\frac{ ? }{ x-1 }$
Since the factor in the denominator is linear, the numerator is an unknown constant $B$$\frac{ A }{ 3x+1 }+\frac{ B }{ x-1 }$
To get the unknown values, set the sum of fractions equal to the original fraction$\frac{ 8x-4 }{ \left( 3x+1 \right) \times \left( x-1 \right) }=\frac{ A }{ 3x+1 }+\frac{ B }{ x-1 }$
Multiply both sides of the equation by $\left( 3x+1 \right) \times \left( x-1 \right)$$8x-4=\left( x-1 \right)A+\left( 3x+1 \right)B$
Simplify the expression$8x-4=Ax-A+3Bx+B$
Use the commutative property to reorder the terms$8x-4=Ax+3Bx-A+B$
Group the powers of the $x$-terms and the constant terms$8x-4=\left( A+3B \right)x+\left( -A+B \right)$
When two polynomials are equal, their corresponding coefficients must be equal$\left\{\begin{array} { l } -4=-A+B \\ 8=A+3B\end{array} \right.$
Solve the system of equations$\left( A, B\right)=\left( 5, 1\right)$
Substitute the given values into the formed partial-fraction decomposition$\frac{ 5 }{ 3x+1 }+\frac{ 1 }{ x-1 }$
Return the partial-fraction decomposition into the expression$2x+3+\frac{ 5 }{ 3x+1 }+\frac{ 1 }{ x-1 }$