Blog: `3/(x^4 + x)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Friday, August 16, 2013

$\displaystyle\frac{{3}}{{{{x}}^{{4}}+{x}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

$\displaystyle\frac{{3}}{{{{x}}^{{4}}+{x}}}$

Let's factorize the denominator,

$\displaystyle{{x}}^{{4}}+{x}={x}{\left({{x}}^{{3}}+{1}\right)}$

$\displaystyle={x}{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{x}+{1}\right)}$

Let $\displaystyle\frac{{3}}{{{{x}}^{{4}}+{x}}}=\frac{{A}}{{x}}+\frac{{B}}{{{x}+{1}}}+\frac{{{C}{x}+{D}}}{{{{x}}^{{2}}-{x}+{1}}}$

$\displaystyle\frac{{3}}{{{{x}}^{{4}}+{x}}}=\frac{{{A}{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{x}+{1}\right)}+{B}{\left({x}\right)}{\left({{x}}^{{2}}-{x}+{1}\right)}+{\left({C}{x}+{D}\right)}{\left({x}\right)}{\left({x}+{1}\right)}}}{{{x}{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{x}+{1}\right)}}}$

$\displaystyle\frac{{3}}{{{{x}}^{{4}}+{x}}}=\frac{{{A}{\left({{x}}^{{3}}-{{x}}^{{2}}+{x}+{{x}}^{{2}}-{x}+{1}\right)}+{B}{\left({{x}}^{{3}}-{{x}}^{{2}}+{x}\right)}+{\left({C}{x}+{D}\right)}{\left({{x}}^{{2}}+{x}\right)}}}{{{x}{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{x}+{1}\right)}}}$

$\displaystyle\frac{{3}}{{{{x}}^{{4}}+{x}}}=\frac{{{A}{\left({{x}}^{{3}}+{1}\right)}+{B}{\left({{x}}^{{3}}-{{x}}^{{2}}+{x}\right)}+{C}{{x}}^{{3}}+{C}{{x}}^{{2}}+{D}{{x}}^{{2}}+{D}{x}}}{{{x}{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{x}+{1}\right)}}}$

$\displaystyle\frac{{3}}{{{{x}}^{{4}}+{x}}}=\frac{{{{x}}^{{3}}{\left({A}+{B}+{C}\right)}+{{x}}^{{2}}{\left(-{B}+{C}+{D}\right)}+{x}{\left({B}+{D}\right)}+{A}}}{{{x}{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{x}+{1}\right)}}}$

$\displaystyle\therefore{3}={{x}}^{{3}}{\left({A}+{B}+{C}\right)}+{{x}}^{{2}}{\left(-{B}+{C}+{D}\right)}+{x}{\left({B}+{D}\right)}+{A}$

equating the coefficients of the like terms,

$\displaystyle{A}+{B}+{C}={0}$ - equation 1

$\displaystyle-{B}+{C}+{D}={0}$ - equation 2

$\displaystyle{B}+{D}={0}$ - equation 3

$\displaystyle{A}={3}$

Plug the value of A in equation 1,

$\displaystyle{3}+{B}+{C}={0}$

$\displaystyle{B}+{C}=-{3}$

$\displaystyle{C}=-{3}-{B}$

Substitute the above expression of C in equation 2,

$\displaystyle-{B}+{\left(-{3}-{B}\right)}+{D}={0}$

$\displaystyle-{B}-{3}-{B}+{D}={0}$

$\displaystyle-{2}{B}+{D}={3}$ - equation 4

Now solve equations 3 and...

$\displaystyle\frac{{3}}{{{{x}}^{{4}}+{x}}}$

Let's factorize the denominator,

$\displaystyle{{x}}^{{4}}+{x}={x}{\left({{x}}^{{3}}+{1}\right)}$

$\displaystyle={x}{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{x}+{1}\right)}$

Let $\displaystyle\frac{{3}}{{{{x}}^{{4}}+{x}}}=\frac{{A}}{{x}}+\frac{{B}}{{{x}+{1}}}+\frac{{{C}{x}+{D}}}{{{{x}}^{{2}}-{x}+{1}}}$

$\displaystyle\therefore{3}={{x}}^{{3}}{\left({A}+{B}+{C}\right)}+{{x}}^{{2}}{\left(-{B}+{C}+{D}\right)}+{x}{\left({B}+{D}\right)}+{A}$

equating the coefficients of the like terms,

$\displaystyle{A}+{B}+{C}={0}$ - equation 1

$\displaystyle-{B}+{C}+{D}={0}$ - equation 2

$\displaystyle{B}+{D}={0}$ - equation 3

$\displaystyle{A}={3}$

Plug the value of A in equation 1,

$\displaystyle{3}+{B}+{C}={0}$

$\displaystyle{B}+{C}=-{3}$

$\displaystyle{C}=-{3}-{B}$

Substitute the above expression of C in equation 2,

$\displaystyle-{B}+{\left(-{3}-{B}\right)}+{D}={0}$

$\displaystyle-{B}-{3}-{B}+{D}={0}$

$\displaystyle-{2}{B}+{D}={3}$ - equation 4

Now solve equations 3 and 4 to get the solutions of B and D,

Subtract equation 3 from equation 4,

$\displaystyle{\left(-{2}{B}+{D}\right)}-{\left({B}+{D}\right)}={3}-{0}$

$\displaystyle-{3}{B}={3}$

$\displaystyle{B}=-{1}$

Plug the value of B in equation 3,

$\displaystyle-{1}+{D}={0}$

$\displaystyle{D}={1}$

Plug the value of A and B in equation 1,

$\displaystyle{3}+{\left(-{1}\right)}+{C}={0}$

$\displaystyle{2}+{C}={0}$

$\displaystyle{C}=-{2}$

$\displaystyle\therefore\frac{{3}}{{{{x}}^{{4}}+{x}}}=\frac{{3}}{{x}}-\frac{{1}}{{{x}+{1}}}+\frac{{-{2}{x}+{1}}}{{{{x}}^{{2}}-{x}+{1}}}$

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Friday, August 16, 2013

$\displaystyle\frac{{3}}{{{{x}}^{{4}}+{x}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

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What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?

Friday, August 16, 2013

Write the partial fraction decomposition of the rational expression. Check your result algebraically.

No comments:

Post a Comment

What is the Exposition, Rising Action, Climax, and Falling Action of &quot;One Thousand Dollars&quot;?

$\displaystyle\frac{{3}}{{{{x}}^{{4}}+{x}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?