$\frac{ ? }{ x+3 }+\frac{ ? }{ x-2 }$
Since the factor in the denominator is linear, the numerator is an unknown constant $A$$\frac{ A }{ x+3 }+\frac{ ? }{ x-2 }$
Since the factor in the denominator is linear, the numerator is an unknown constant $B$$\frac{ A }{ x+3 }+\frac{ B }{ x-2 }$
To get the unknown values, set the sum of fractions equal to the original fraction$\frac{ 6x-2 }{ \left( x+3 \right) \times \left( x-2 \right) }=\frac{ A }{ x+3 }+\frac{ B }{ x-2 }$
Multiply both sides of the equation by $\left( x+3 \right) \times \left( x-2 \right)$$6x-2=\left( x-2 \right)A+\left( x+3 \right)B$
Simplify the expression$6x-2=Ax-2A+Bx+3B$
Use the commutative property to reorder the terms$6x-2=Ax+Bx-2A+3B$
Group the powers of the $x$-terms and the constant terms$6x-2=\left( A+B \right)x+\left( -2A+3B \right)$
When two polynomials are equal, their corresponding coefficients must be equal$\left\{\begin{array} { l } -2=-2A+3B \\ 6=A+B\end{array} \right.$
Solve the system of equations$\left( A, B\right)=\left( 4, 2\right)$
Substitute the given values into the formed partial-fraction decomposition$\frac{ 4 }{ x+3 }+\frac{ 2 }{ x-2 }$