# Solve for: (6x^3+5x^2-7)/(3x^2-2x-1)

## Expression: $\frac{ 6{x}^{3}+5{x}^{2}-7 }{ 3{x}^{2}-2x-1 }$

Since the numerator needs to have a lower degree than the denominator, divide the numerator by the denominator using long division

$2x+3+\frac{ 8x-4 }{ 3{x}^{2}-2x-1 }$

Write $-2x$ as a difference

$2x+3+\frac{ 8x-4 }{ 3{x}^{2}+x-3x-1 }$

Factor out $x$ from the expression

$2x+3+\frac{ 8x-4 }{ x \times \left( 3x+1 \right)-3x-1 }$

Factor out the negative sign from the expression

$2x+3+\frac{ 8x-4 }{ x \times \left( 3x+1 \right)-\left( 3x+1 \right) }$

Factor out $3x+1$ from the expression

$2x+3+\frac{ 8x-4 }{ \left( 3x+1 \right) \times \left( x-1 \right) }$

Prepare for partial-fraction decomposition by isolating the fraction

$\frac{ 8x-4 }{ \left( 3x+1 \right) \times \left( x-1 \right) }$

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

$\frac{ ? }{ 3x+1 }+\frac{ ? }{ x-1 }$

Since the factor in the denominator is linear, the numerator is an unknown constant $A$

$\frac{ A }{ 3x+1 }+\frac{ ? }{ x-1 }$

Since the factor in the denominator is linear, the numerator is an unknown constant $B$

$\frac{ A }{ 3x+1 }+\frac{ B }{ x-1 }$

To get the unknown values, set the sum of fractions equal to the original fraction

$\frac{ 8x-4 }{ \left( 3x+1 \right) \times \left( x-1 \right) }=\frac{ A }{ 3x+1 }+\frac{ B }{ x-1 }$

Multiply both sides of the equation by $\left( 3x+1 \right) \times \left( x-1 \right)$

$8x-4=\left( x-1 \right)A+\left( 3x+1 \right)B$

Simplify the expression

$8x-4=Ax-A+3Bx+B$

Use the commutative property to reorder the terms

$8x-4=Ax+3Bx-A+B$

Group the powers of the $x$-terms and the constant terms

$8x-4=\left( A+3B \right)x+\left( -A+B \right)$

When two polynomials are equal, their corresponding coefficients must be equal

$\left\{\begin{array} { l } -4=-A+B \\ 8=A+3B\end{array} \right.$

Solve the system of equations

$\left( A, B\right)=\left( 5, 1\right)$

Substitute the given values into the formed partial-fraction decomposition

$\frac{ 5 }{ 3x+1 }+\frac{ 1 }{ x-1 }$

Return the partial-fraction decomposition into the expression

$2x+3+\frac{ 5 }{ 3x+1 }+\frac{ 1 }{ x-1 }$

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