Question

To find: The partial fraction decomposition for the rational expression \frac{11-2x}{x^{2}-8x+16}.

Factors and multiples
ANSWERED
asked 2021-08-15
To find:
The partial fraction decomposition for the rational expression \(\displaystyle{\frac{{{11}-{2}{x}}}{{{x}^{{{2}}}-{8}{x}+{16}}}}\).

Answers (1)

2021-08-16

Steps to solve:
Partial fraction decomposition of \(\displaystyle{\frac{{{f{{\left({x}\right)}}}}}{{{g{{\left({x}\right)}}}}}}\): To form a partial fraction decomposition of a rational expression, follow these steps.
1. If \(\displaystyle{\frac{{{f{{\left({x}\right)}}}}}{{{g{{\left({x}\right)}}}}}}\) is not a proper fraction (a fraction with the numerator of lesser degree than the denominator), divide f(x) by g(x).
2. Factor the denominator g(x) completely into factors of the form \(\displaystyle{\left({a}{x}+{b}\right)}^{{{m}}}\) or \(\displaystyle{\left({c}{x}^{{{2}}}+{\left.{d}{x}\right.}+{e}\right)}^{{{n}}}\), where \(\displaystyle{c}{x}^{{{2}}}+{\left.{d}{x}\right.}+{e}\) is irreducible and m and n are positive integers.
3. For each repeated linear factor \(\displaystyle{\left({a}{x}+{b}\right)}^{{{m}}}\), the decomposition must include the terms \(\displaystyle{\frac{{{A}_{{{1}}}}}{{{a}{x}+{b}}}}+{\frac{{{A}_{{{2}}}}}{{{\left({a}{x}+{b}\right)}^{{{2}}}}}}+\cdots+{\frac{{{A}_{{{m}}}}}{{{\left({a}{x}+{b}\right)}^{{{m}}}}}}\).
4. Use algebraic techniques to solve for the constants in the numerators or the decomposition.
Consider the rational expression. \(\displaystyle{\frac{{{11}-{2}{x}}}{{{x}{\left\lbrace{2}\right\rbrace}-{8}{x}+{16}}}}\)
The factor of \(\displaystyle{x}^{{{2}}}-{8}{x}+{16}={\left({x}-{4}\right)}{\left({x}-{4}\right)}={\left({x}-{4}\right)}^{{{2}}}\)
As the denominator is a repeated linear factor \(\displaystyle{\left({x}-{4}\right)}^{{{2}}}\), the decomposition must include the terms \(\displaystyle{\frac{{{A}}}{{{\left({x}-{4}\right)}}}}+{\frac{{{B}}}{{{\left({x}-{4}\right)}^{{{2}}}}}}\).
The rational expression becomes \(\displaystyle{\frac{{{11}-{2}{x}}}{{{x}^{{{2}}}-{8}{x}+{16}}}}={\frac{{{A}}}{{{\left({x}-{4}\right)}}}}+{\frac{{{B}}}{{{\left({x}-{4}\right)}^{{{2}}}}}}\rightarrow{\left({1}\right)}\)
Taking least common multiples on both the sides of the denominator,
\(\displaystyle{\frac{{{11}-{2}{x}}}{{{x}^{{{2}}}-{8}{x}+{16}}}}={\frac{{{\left({x}-{4}\right)}{A}+{B}}}{{{\left({x}-{4}\right)}^{{{2}}}}}}\)
\(11-2x=(x-4)A+B\rightarrow\)(2)
Put \(\displaystyle{x}={4}\) to get the value of B from equation(2).
\(\displaystyle{11}-{2}{\left({4}\right)}={\left({4}-{4}\right)}{A}+{B}\)
\(\displaystyle{3}={B}\)
Put \(\displaystyle{x}={0}\) to get the value of A from equation (2)
\(\displaystyle{11}-{2}{\left({0}\right)}={\left({0}-{4}\right)}{A}+{B}\)
\(\displaystyle{11}=-{4}{A}+{B}\rightarrow{\left({3}\right)}\)
Now substitute the value of B in equation (3)
\(\displaystyle{11}=-{4}{A}+{\left({3}\right)}\)
\(\displaystyle-{4}{A}={8}\)
\(\displaystyle{A}=-{2}\)
Substituting the values of A and B in equation (1),
\(\displaystyle{\frac{{{11}-{2}{x}}}{{{x}^{{{2}}}-{8}{x}+{16}}}}={\frac{{-{2}}}{{{\left({x}-{4}\right)}}}}+{\frac{{{3}}}{{{\left({x}-{4}\right)}^{{{2}}}}}}\)
Therefore, the partial fraction decomposition of the rational expression \(\displaystyle{\frac{{{11}-{2}{x}}}{{{x}^{{{2}}}-{8}{x}+{16}}}}\) is \(\displaystyle{\frac{{{3}}}{{{\left({x}-{4}\right)}^{{{2}}}}}}-{\frac{{{2}}}{{{\left({x}-{4}\right)}}}}\).

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