# Calculate: (6x-2)/((x+3) * (x-2))

## Expression: $\frac{ 6x-2 }{ \left( x+3 \right) \times \left( x-2 \right) }$

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

$\frac{ ? }{ x+3 }+\frac{ ? }{ x-2 }$

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

$\frac{ 4 }{ x+3 }+\frac{ 2 }{ x-2 }$

Random Posts
Random Articles