Blog: `(x^2 + 5)/((x + 1)(x^2 - 2x + 3))` Write the partial fraction decomposition of the rational expression. Check your result algebraically.

Sunday, June 30, 2013

$\displaystyle\frac{{{{x}}^{{2}}+{5}}}{{{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}}}$ Write the partial fraction decomposition of the rational expression. Check your result algebraically.

$\displaystyle\frac{{{{x}}^{{2}}+{5}}}{{{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}}}$

Let $\displaystyle\frac{{{{x}}^{{2}}+{5}}}{{{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}}}=\frac{{A}}{{{x}+{1}}}+\frac{{{B}{x}+{C}}}{{{{x}}^{{2}}-{2}{x}+{3}}}$

$\displaystyle\frac{{{{x}}^{{2}}+{5}}}{{{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}}}=\frac{{{A}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}+{\left({B}{x}+{C}\right)}{\left({x}+{1}\right)}}}{{{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}}}$

$\displaystyle\frac{{{{x}}^{{2}}+{5}}}{{{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}}}=\frac{{{A}{{x}}^{{2}}-{2}{A}{x}+{3}{A}+{B}{{x}}^{{2}}+{B}{x}+{C}{x}+{C}}}{{{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}}}$

$\displaystyle\therefore{\left({{x}}^{{2}}+{5}\right)}={A}{{x}}^{{2}}-{2}{A}{x}+{3}{A}+{B}{{x}}^{{2}}+{B}{x}+{C}{x}+{C}$

$\displaystyle{{x}}^{{2}}+{5}={\left({A}+{B}\right)}{{x}}^{{2}}+{\left(-{2}{A}+{B}+{C}\right)}{x}+{3}{A}+{C}$

equating the coefficients of the like terms,

$\displaystyle{A}+{B}={1}$

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

$\displaystyle{3}{A}+{C}={5}$

Now let's solve the above three equations to find the values of A,B and C,

Express C in terms of A from the third equation,

$\displaystyle{C}={5}-{3}{A}$

Substitute the above expression of C in second equation,

$\displaystyle-{2}{A}+{B}+{5}-{3}{A}={0}$

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

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

Now subtract the first equation from the above equation,

$\displaystyle{\left(-{5}{A}+{B}\right)}-{\left({A}+{B}\right)}=-{5}-{1}$

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

$\displaystyle{A}={1}$

Plug the value of A in the first and...

$\displaystyle\frac{{{{x}}^{{2}}+{5}}}{{{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}}}$

Let $\displaystyle\frac{{{{x}}^{{2}}+{5}}}{{{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}}}=\frac{{A}}{{{x}+{1}}}+\frac{{{B}{x}+{C}}}{{{{x}}^{{2}}-{2}{x}+{3}}}$

$\displaystyle\therefore{\left({{x}}^{{2}}+{5}\right)}={A}{{x}}^{{2}}-{2}{A}{x}+{3}{A}+{B}{{x}}^{{2}}+{B}{x}+{C}{x}+{C}$

$\displaystyle{{x}}^{{2}}+{5}={\left({A}+{B}\right)}{{x}}^{{2}}+{\left(-{2}{A}+{B}+{C}\right)}{x}+{3}{A}+{C}$

equating the coefficients of the like terms,

$\displaystyle{A}+{B}={1}$

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

$\displaystyle{3}{A}+{C}={5}$

Now let's solve the above three equations to find the values of A,B and C,

Express C in terms of A from the third equation,

$\displaystyle{C}={5}-{3}{A}$

Substitute the above expression of C in second equation,

$\displaystyle-{2}{A}+{B}+{5}-{3}{A}={0}$

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

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

Now subtract the first equation from the above equation,

$\displaystyle{\left(-{5}{A}+{B}\right)}-{\left({A}+{B}\right)}=-{5}-{1}$

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

$\displaystyle{A}={1}$

Plug the value of A in the first and third equation to get the values of B and C,

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

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

$\displaystyle{B}={0}$

$\displaystyle{3}{\left({1}\right)}+{C}={5}$

$\displaystyle{C}={5}-{3}$

$\displaystyle{C}={2}$

$\displaystyle\therefore\frac{{{{x}}^{{2}}+{5}}}{{{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}}}=\frac{{1}}{{{x}+{1}}}+\frac{{2}}{{{{x}}^{{2}}-{2}{x}+{3}}}$

Now let's check it algebraically,

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

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

$\displaystyle=\frac{{{{x}}^{{2}}+{5}}}{{{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}}}$

Hence it is verified.

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Sunday, June 30, 2013

$\displaystyle\frac{{{{x}}^{{2}}+{5}}}{{{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}}}$ 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"?

Sunday, June 30, 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;?

$\displaystyle\frac{{{{x}}^{{2}}+{5}}}{{{\left({x}+{1}\right)}{\left({{x}}^{{2}}-{2}{x}+{3}\right)}}}$ 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"?