$\frac{ 7x+6+5{x}^{2} }{ x \times \left( x-2+{x}^{2} \right) }$
Use the commutative property to reorder the terms$\frac{ 7x+6+5{x}^{2} }{ x \times \left( {x}^{2}+x-2 \right) }$
Write $x$ as a difference$\frac{ 7x+6+5{x}^{2} }{ x \times \left( {x}^{2}+2x-x-2 \right) }$
Factor out $x$ from the expression$\frac{ 7x+6+5{x}^{2} }{ x \times \left( x \times \left( x+2 \right)-x-2 \right) }$
Factor out the negative sign from the expression$\frac{ 7x+6+5{x}^{2} }{ x \times \left( x \times \left( x+2 \right)-\left( x+2 \right) \right) }$
Factor out $x+2$ from the expression$\frac{ 7x+6+5{x}^{2} }{ x \times \left( x+2 \right) \times \left( x-1 \right) }$
For each factor in the denominator, write a new fraction using the factors as new denominators. The numerators are unknown values$\frac{ ? }{ x }+\frac{ ? }{ x+2 }+\frac{ ? }{ x-1 }$
Since the factor in the denominator is linear, the numerator is an unknown constant $A$$\frac{ A }{ x }+\frac{ ? }{ x+2 }+\frac{ ? }{ x-1 }$
Since the factor in the denominator is linear, the numerator is an unknown constant $B$$\frac{ A }{ x }+\frac{ B }{ x+2 }+\frac{ ? }{ x-1 }$
Since the factor in the denominator is linear, the numerator is an unknown constant $C$$\frac{ A }{ x }+\frac{ B }{ x+2 }+\frac{ C }{ x-1 }$
To get the unknown values, set the sum of fractions equal to the original fraction$\frac{ 7x+6+5{x}^{2} }{ x \times \left( x+2 \right) \times \left( x-1 \right) }=\frac{ A }{ x }+\frac{ B }{ x+2 }+\frac{ C }{ x-1 }$
Multiply both sides of the equation by $x \times \left( x+2 \right) \times \left( x-1 \right)$$7x+6+5{x}^{2}=\left( x+2 \right) \times \left( x-1 \right)A+x \times \left( x-1 \right)B+x \times \left( x+2 \right)C$
Simplify the expression$7x+6+5{x}^{2}=A{x}^{2}+Ax-2A+B{x}^{2}-Bx+C{x}^{2}+2Cx$
Use the commutative property to reorder the terms$7x+6+5{x}^{2}=A{x}^{2}+B{x}^{2}+C{x}^{2}+Ax-Bx+2Cx-2A$
Group the powers of the $x$-terms and the constant terms$7x+6+5{x}^{2}=\left( A+B+C \right){x}^{2}+\left( A-B+2C \right)x-2A$
When two polynomials are equal, their corresponding coefficients must be equal$\left\{\begin{array} { l } 6=-2A \\ 7=A-B+2C \\ 5=A+B+C\end{array} \right.$
Solve the system of equations$\left( A, B, C\right)=\left( -3, 2, 6\right)$
Substitute the given values into the formed partial-fraction decomposition$\frac{ -3 }{ x }+\frac{ 2 }{ x+2 }+\frac{ 6 }{ x-1 }$
Use $\frac{ -a }{ b }=\frac{ a }{ -b }=-\frac{ a }{ b }$ to rewrite the fraction$-\frac{ 3 }{ x }+\frac{ 2 }{ x+2 }+\frac{ 6 }{ x-1 }$