$\frac{ ? }{ x-1 }+\frac{ ? }{ {x}^{2}+x+1 }$
Since the factor in the denominator is linear, the numerator is an unknown constant $A$$\frac{ A }{ x-1 }+\frac{ ? }{ {x}^{2}+x+1 }$
Since the factor in the denominator is quadratic, the numerator is an unknown linear expression $Bx+C$$\frac{ A }{ x-1 }+\frac{ Bx+C }{ {x}^{2}+x+1 }$
To get the unknown values, set the sum of fractions equal to the original fraction$\frac{ 10x+2 }{ \left( x-1 \right) \times \left( {x}^{2}+x+1 \right) }=\frac{ A }{ x-1 }+\frac{ Bx+C }{ {x}^{2}+x+1 }$
Multiply both sides of the equation by $\left( x-1 \right) \times \left( {x}^{2}+x+1 \right)$$10x+2=\left( {x}^{2}+x+1 \right)A+\left( x-1 \right) \times \left( Bx+C \right)$
Simplify the expression$10x+2=A{x}^{2}+Ax+A+B{x}^{2}+Cx-Bx-C$
Use the commutative property to reorder the terms$10x+2=A{x}^{2}+B{x}^{2}+Ax+Cx-Bx+A-C$
Group the powers of the $x$-terms and the constant terms$10x+2=\left( A+B \right){x}^{2}+\left( A+C-B \right)x+\left( A-C \right)$
When two polynomials are equal, their corresponding coefficients must be equal$\left\{\begin{array} { l } 2=A-C \\ 10=A+C-B \\ 0=A+B\end{array} \right.$
Solve the system of equations$\left( A, B, C\right)=\left( 4, -4, 2\right)$
Substitute the given values into the formed partial-fraction decomposition$\frac{ 4 }{ x-1 }+\frac{ -4x+2 }{ {x}^{2}+x+1 }$