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

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

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