Solve for: (10x+2)/((x-1) * (x^2+x+1))

Expression: $\frac{ 10x+2 }{ \left( x-1 \right) \times \left( {x}^{2}+x+1 \right) }$

For each factor in the denominator, write a new fraction using the factors as new denominators. The numerators are unknown values

$\frac{ ? }{ x-1 }+\frac{ ? }{ {x}^{2}+x+1 }$

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

$\frac{ 4 }{ x-1 }+\frac{ -4x+2 }{ {x}^{2}+x+1 }$