Blog: `(x^3 + 2x^2 - x + 1)/(x^2 + 3x - 4)` Write the partial fraction decomposition of the improper rational expression.

Thursday, March 7, 2013

$\displaystyle\frac{{{{x}}^{{3}}+{2}{{x}}^{{2}}-{x}+{1}}}{{{{x}}^{{2}}+{3}{x}-{4}}}$ Write the partial fraction decomposition of the improper rational expression.

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

Since the rational expression is an improper expression , we have to express the expression as a sum of simpler fractions with the degree of the polynomial in the numerator less than the degree of the polynomial in the denominator.

Dividing using the long division yields,

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

Polynomials do not completely divide , so we have to continue with the partial fractions of the remainder expression,

Let's factorize the denominator of the remainder...

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

Dividing using the long division yields,

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

Polynomials do not completely divide , so we have to continue with the partial fractions of the remainder expression,

Let's factorize the denominator of the remainder fraction,

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

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

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

Let $\displaystyle\frac{{{6}{x}-{3}}}{{{{x}}^{{2}}+{3}{x}-{4}}}=\frac{{A}}{{{x}-{1}}}+\frac{{B}}{{{x}+{4}}}$

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

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

$\displaystyle\therefore{\left({6}{x}-{3}\right)}={A}{x}+{4}{A}+{B}{x}-{B}$

$\displaystyle{6}{x}-{3}={x}{\left({A}+{B}\right)}+{4}{A}-{B}$

equating the coefficients of the like terms,

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

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

Now we have to solve the above equations to get the solutions of A and B,

Adding the equation 1 and 2 yields,

$\displaystyle{A}+{4}{A}={6}+{\left(-{3}\right)}$

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

$\displaystyle{A}=\frac{{3}}{{5}}$

Plug the value of A in equation 1 ,

$\displaystyle\frac{{3}}{{5}}+{B}={6}$

$\displaystyle{B}={6}-\frac{{3}}{{5}}$

$\displaystyle{B}=\frac{{27}}{{5}}$

$\displaystyle\frac{{{6}{x}-{3}}}{{{{x}}^{{2}}+{3}{x}-{4}}}=\frac{{3}}{{{5}{\left({x}-{1}\right)}}}+\frac{{27}}{{{5}{\left({x}+{4}\right)}}}$

$\displaystyle\therefore\frac{{{{x}}^{{3}}+{2}{{x}}^{{2}}-{x}+{1}}}{{{{x}}^{{2}}+{3}{x}-{4}}}={x}-{1}+\frac{{3}}{{{5}{\left({x}-{1}\right)}}}+\frac{{27}}{{{5}{\left({x}+{4}\right)}}}$

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Thursday, March 7, 2013

$\displaystyle\frac{{{{x}}^{{3}}+{2}{{x}}^{{2}}-{x}+{1}}}{{{{x}}^{{2}}+{3}{x}-{4}}}$ Write the partial fraction decomposition of the improper rational expression.

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

Thursday, March 7, 2013

Write the partial fraction decomposition of the improper rational expression.

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Post a Comment

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

$\displaystyle\frac{{{{x}}^{{3}}+{2}{{x}}^{{2}}-{x}+{1}}}{{{{x}}^{{2}}+{3}{x}-{4}}}$ Write the partial fraction decomposition of the improper rational expression.

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