To find: The partial fraction decomposition for the rational expression \frac{11-2x}{x^{2}-8x+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 ${\displaystyle {(ax+b)}^m}$ or ${\displaystyle {(c{x}^2+dx+e)}^n}$, where ${\displaystyle c{x}^2+dx+e}$ is irreducible and m and n are positive integers.
3. For each repeated linear factor ${\displaystyle {(ax+b)}^m}$, the decomposition must include the terms ${\displaystyle \frac{A_1}{ax+b}+\frac{A_2}{{(ax+b)}^2}+\cdots +\frac{A_m}{{(ax+b)}^m}}$.
4. Use algebraic techniques to solve for the constants in the numerators or the decomposition.
Consider the rational expression. ${\displaystyle \frac{11-2x}{x\left\{2\right\}-8x+16}}$
The factor of ${\displaystyle x^2-8x+16=(x-4)(x-4)={(x-4)}^2}$
As the denominator is a repeated linear factor ${\displaystyle {(x-4)}^2}$, the decomposition must include the terms ${\displaystyle \frac{A}{(x-4)}+\frac{B}{{(x-4)}^2}}$.
The rational expression becomes ${\displaystyle \frac{11-2x}{x^2-8x+16}=\frac{A}{(x-4)}+\frac{B}{{(x-4)}^2}\to \left(1\right)}$
Taking least common multiples on both the sides of the denominator,
${\displaystyle \frac{11-2x}{x^2-8x+16}=\frac{(x-4)A+B}{{(x-4)}^2}}$
$11-2x=(x-4)A+B\to$(2)
Put ${\displaystyle x=4}$ to get the value of B from equation(2).
${\displaystyle 11-2\left(4\right)=(4-4)A+B}$
${\displaystyle 3=B}$
Put ${\displaystyle x=0}$ to get the value of A from equation (2)
${\displaystyle 11-2\left(0\right)=(0-4)A+B}$
${\displaystyle 11=-4A+B\to \left(3\right)}$
Now substitute the value of B in equation (3)
${\displaystyle 11=-4A+\left(3\right)}$
${\displaystyle -4A=8}$
${\displaystyle A=-2}$
Substituting the values of A and B in equation (1),