Blog: `x^2/(x^4 - 2x^2 - 8)` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Thursday, August 18, 2016

$\displaystyle\frac{{{x}}^{{2}}}{{{{x}}^{{4}}-{2}{{x}}^{{2}}-{8}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

$\displaystyle\frac{{{x}}^{{2}}}{{{{x}}^{{4}}-{2}{{x}}^{{2}}-{8}}}$

Let's factorize the denominator,

$\displaystyle{{x}}^{{4}}-{2}{{x}}^{{2}}-{8}={{x}}^{{4}}-{4}{{x}}^{{2}}+{2}{{x}}^{{2}}-{8}$

$\displaystyle={{x}}^{{2}}{\left({{x}}^{{2}}-{4}\right)}+{2}{\left({{x}}^{{2}}-{4}\right)}$

$\displaystyle={\left({{x}}^{{2}}+{2}\right)}{\left({{x}}^{{2}}-{4}\right)}$

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

$\displaystyle\therefore\frac{{{x}}^{{2}}}{{{{x}}^{{4}}-{2}{{x}}^{{2}}-{8}}}=\frac{{{x}}^{{2}}}{{{\left({x}+{2}\right)}{\left({x}-{2}\right)}{\left({{x}}^{{2}}+{2}\right)}}}$

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

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

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

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

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

Now,

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

Equating the coefficients of like terms,

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

$\displaystyle-{2}{A}+{2}{B}+{D}={1}$ ------ equation 2

$\displaystyle{2}{A}+{2}{B}-{4}{C}={0}$ ------ equation 3

$\displaystyle-{4}{A}+{4}{B}-{4}{D}={0}$ ----- equation 4

Now we have to solve the above four equations to find the solutions of A,B,C and D.

Rewrite equation 1 ,

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

Substitute above in equation 3,

$\displaystyle{2}{\left({A}+{B}\right)}-{4}{C}={0}$

$\displaystyle{2}{\left(-{C}\right)}-{4}{C}={0}$

$\displaystyle-{2}{C}-{4}{C}={0}$

$\displaystyle-{6}{C}={0}$

$\displaystyle{C}={0}$

Rewrite equation...

$\displaystyle\frac{{{x}}^{{2}}}{{{{x}}^{{4}}-{2}{{x}}^{{2}}-{8}}}$

Let's factorize the denominator,

$\displaystyle{{x}}^{{4}}-{2}{{x}}^{{2}}-{8}={{x}}^{{4}}-{4}{{x}}^{{2}}+{2}{{x}}^{{2}}-{8}$

$\displaystyle={{x}}^{{2}}{\left({{x}}^{{2}}-{4}\right)}+{2}{\left({{x}}^{{2}}-{4}\right)}$

$\displaystyle={\left({{x}}^{{2}}+{2}\right)}{\left({{x}}^{{2}}-{4}\right)}$

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

$\displaystyle\therefore\frac{{{x}}^{{2}}}{{{{x}}^{{4}}-{2}{{x}}^{{2}}-{8}}}=\frac{{{x}}^{{2}}}{{{\left({x}+{2}\right)}{\left({x}-{2}\right)}{\left({{x}}^{{2}}+{2}\right)}}}$

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

Now,

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

Equating the coefficients of like terms,

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

$\displaystyle-{2}{A}+{2}{B}+{D}={1}$ ------ equation 2

$\displaystyle{2}{A}+{2}{B}-{4}{C}={0}$ ------ equation 3

$\displaystyle-{4}{A}+{4}{B}-{4}{D}={0}$ ----- equation 4

Now we have to solve the above four equations to find the solutions of A,B,C and D.

Rewrite equation 1 ,

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

Substitute above in equation 3,

$\displaystyle{2}{\left({A}+{B}\right)}-{4}{C}={0}$

$\displaystyle{2}{\left(-{C}\right)}-{4}{C}={0}$

$\displaystyle-{2}{C}-{4}{C}={0}$

$\displaystyle-{6}{C}={0}$

$\displaystyle{C}={0}$

Rewrite equation 2 as,

$\displaystyle-{2}{A}+{2}{B}={1}-{D}$

Substitute the above in equation 4,

$\displaystyle-{4}{A}+{4}{B}-{4}{D}={0}$

$\displaystyle{2}{\left(-{2}{A}+{2}{B}\right)}-{4}{D}={0}$

$\displaystyle{2}{\left({1}-{D}\right)}-{4}{D}={0}$

$\displaystyle{2}-{2}{D}-{4}{D}={0}$

$\displaystyle{2}-{6}{D}={0}$

$\displaystyle{D}=\frac{{2}}{{6}}$

$\displaystyle{D}=\frac{{1}}{{3}}$

Plug the values of C in equation 3 ,

$\displaystyle{2}{A}+{2}{B}-{4}{\left({0}\right)}={0}$

$\displaystyle{2}{A}+{2}{B}={0}$

$\displaystyle{2}{\left({A}+{B}\right)}={0}$

$\displaystyle{A}+{B}={0}$ ------ equation 5

Plug the value of D in equation 4,

$\displaystyle-{4}{A}+{4}{B}-{4}{\left(\frac{{1}}{{3}}\right)}={0}$

$\displaystyle-{4}{A}+{4}{B}=\frac{{4}}{{3}}$

$\displaystyle{4}{\left(-{A}+{B}\right)}=\frac{{4}}{{3}}$

$\displaystyle-{A}+{B}=\frac{{1}}{{3}}$ ---- equation 6

Solve equations 5 and 6 to find the solutions of A and B,

Add the equations 5 and 6.

$\displaystyle{2}{B}=\frac{{1}}{{3}}$

$\displaystyle{B}=\frac{{1}}{{6}}$

Plug the value of B in equation 5,

$\displaystyle{A}+\frac{{1}}{{6}}={0}$

$\displaystyle{A}=-\frac{{1}}{{6}}$

$\displaystyle\therefore\frac{{{x}}^{{2}}}{{{{x}}^{{4}}-{2}{{x}}^{{2}}-{8}}}=\frac{{-\frac{{1}}{{6}}}}{{{x}+{2}}}+\frac{{\frac{{1}}{{6}}}}{{{x}-{2}}}+\frac{{{0}\cdot{x}+\frac{{1}}{{3}}}}{{{{x}}^{{2}}+{2}}}$

$\displaystyle\frac{{{x}}^{{2}}}{{{{x}}^{{4}}-{2}{{x}}^{{2}}-{8}}}=-\frac{{1}}{{{6}{\left({x}+{2}\right)}}}+\frac{{1}}{{{6}{\left({x}-{2}\right)}}}+\frac{{1}}{{{3}{\left({{x}}^{{2}}+{2}\right)}}}$

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Thursday, August 18, 2016

$\displaystyle\frac{{{x}}^{{2}}}{{{{x}}^{{4}}-{2}{{x}}^{{2}}-{8}}}$ 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"?

Thursday, August 18, 2016

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;?

$\displaystyle\frac{{{x}}^{{2}}}{{{{x}}^{{4}}-{2}{{x}}^{{2}}-{8}}}$ 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"?