# Solve for: (x^2)/((x-4)^2 * (x+3))

## Expression: $\frac{ {x}^{2} }{ {\left( x-4 \right)}^{2} \times \left( x+3 \right) }$

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

$\frac{ ? }{ {\left( x-4 \right)}^{2} }+\frac{ ? }{ x+3 }$

Since the factor $x-4$ is raised to the power of $2$, it is a repeating factor, so include each power from $1$ to $2$ as denominators of separate fractions

$\frac{ ? }{ x-4 }+\frac{ ? }{ {\left( x-4 \right)}^{2} }+\frac{ ? }{ x+3 }$

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

$\frac{ A }{ x-4 }+\frac{ ? }{ {\left( x-4 \right)}^{2} }+\frac{ ? }{ x+3 }$

Since the denominator is a linear factor raised to a power, the numerator is an unknown constant $B$

$\frac{ A }{ x-4 }+\frac{ B }{ {\left( x-4 \right)}^{2} }+\frac{ ? }{ x+3 }$

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

$\frac{ A }{ x-4 }+\frac{ B }{ {\left( x-4 \right)}^{2} }+\frac{ C }{ x+3 }$

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

$\frac{ {x}^{2} }{ {\left( x-4 \right)}^{2} \times \left( x+3 \right) }=\frac{ A }{ x-4 }+\frac{ B }{ {\left( x-4 \right)}^{2} }+\frac{ C }{ x+3 }$

Multiply both sides of the equation by ${\left( x-4 \right)}^{2} \times \left( x+3 \right)$

${x}^{2}=\left( x-4 \right) \times \left( x+3 \right)A+\left( x+3 \right)B+{\left( x-4 \right)}^{2} \times C$

Simplify the expression

${x}^{2}=A{x}^{2}-Ax-12A+Bx+3B+C{x}^{2}-8Cx+16C$

Use the commutative property to reorder the terms

${x}^{2}=A{x}^{2}+C{x}^{2}-Ax+Bx-8Cx-12A+3B+16C$

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

${x}^{2}=\left( A+C \right){x}^{2}+\left( -A+B-8C \right)x+\left( -12A+3B+16C \right)$

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

$\left\{\begin{array} { l } 0=-12A+3B+16C \\ 0=-A+B-8C \\ 1=A+C\end{array} \right.$

Solve the system of equations

$\left( A, B, C\right)=\left( \frac{ 40 }{ 49 }, \frac{ 16 }{ 7 }, \frac{ 9 }{ 49 }\right)$

Substitute the given values into the formed partial-fraction decomposition

$\frac{ \frac{ 40 }{ 49 } }{ x-4 }+\frac{ \frac{ 16 }{ 7 } }{ {\left( x-4 \right)}^{2} }+\frac{ \frac{ 9 }{ 49 } }{ x+3 }$

Simplify the complex fraction

$\frac{ 40 }{ 49\left( x-4 \right) }+\frac{ \frac{ 16 }{ 7 } }{ {\left( x-4 \right)}^{2} }+\frac{ \frac{ 9 }{ 49 } }{ x+3 }$

Simplify the complex fraction

$\frac{ 40 }{ 49\left( x-4 \right) }+\frac{ 16 }{ 7{\left( x-4 \right)}^{2} }+\frac{ \frac{ 9 }{ 49 } }{ x+3 }$

Simplify the complex fraction

$\frac{ 40 }{ 49\left( x-4 \right) }+\frac{ 16 }{ 7{\left( x-4 \right)}^{2} }+\frac{ 9 }{ 49\left( x+3 \right) }$

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