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

Factors and multiples
ANSWERED To find:
The partial fraction decomposition for the rational expression $$\displaystyle{\frac{{{11}-{2}{x}}}{{{x}^{{{2}}}-{8}{x}+{16}}}}$$. 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 $$\displaystyle{\frac{{{3}}}{{{\left({x}-{4}\right)}^{{{2}}}}}}-{\frac{{{2}}}{{{\left({x}-{4}\right)}}}}$$.