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2021-11-18

To calculate: The partial decomposition of $\frac{{x}^{2}-2x-21}{{x}^{3}+7x}$

Cherry McCormick

Calculation:
Consider the provided expression:
$\frac{{x}^{2}-2x-21}{{x}^{3}+7x}$
Here, $f\left(x\right)=x?-2x-21\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}g\left(x\right)={x}^{\circ }+Tx$
Factorize $g\left(x\right):$
$g\left(x\right)={x}^{3}+7x$
$=x\left({x}^{2}+7\right)$
So, the expression can be written as:
$\frac{{x}^{2}-2x-21}{{x}^{3}+7x}=\frac{{x}^{2}-2x-21}{x\left({x}^{3}+7x\right)}$
Here, in g (x), there are two factors, one linear and one quadratic. So, the expression can be decomposed as:
$\frac{{x}^{2}-2x-21}{x\left({x}^{3}+7x\right)}=\frac{A}{x}+\frac{Bx+C}{{x}^{2}+7}$
Here, Least Common Divisor is $x\left({x}^{2}+7\right)$
Multiply both sides of (1) by the Least Common Divisor to clear fractions:
$x\left({x}^{2}+7\right)\left[\frac{{x}^{2}-2x-21}{x\left({x}^{2}+7x\right)}\right]=x\left({x}^{2}+7\right)\left[\frac{-A}{x}+\frac{Bx+C}{{x}^{2}+7x}\right]$
simplify to obtain:
${x}^{2}+2x-21=A\left({x}^{2}+7\right)+\left(Bx+C\right)x$
${x}^{2}+2x-21=A{x}^{2}+7+B{x}^{2}+Cx$
${x}^{2}+2x-21=\left(A+C\right){x}^{2}+Cx+7A$
Compare the coefficients of x,x? and constant terms:
$A+B=1,C=-2\text{nd7A=-2}nd7A=-21$
Solve for A the equation $7A=—21$
$7A=-21$
$A=-3$
Substitute -3 for A in $A+B=1$ and simplify for B:
$\left(-3\right)+B=1$
$B=4$
Substitute the obtained values of A, B and Cin (1):

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