[math] \newcommand{\ex}[1]{\item } \newcommand{\sx}{\item} \newcommand{\x}{\sx} \newcommand{\sxlab}[1]{} \newcommand{\xlab}{\sxlab} \newcommand{\prov}[1] {\quad #1} \newcommand{\provx}[1] {\quad \mbox{#1}} \newcommand{\intext}[1]{\quad \mbox{#1} \quad} \newcommand{\R}{\mathrm{\bf R}} \newcommand{\Q}{\mathrm{\bf Q}} \newcommand{\Z}{\mathrm{\bf Z}} \newcommand{\C}{\mathrm{\bf C}} \newcommand{\dt}{\textbf} \newcommand{\goesto}{\rightarrow} \newcommand{\ddxof}[1]{\frac{d #1}{d x}} \newcommand{\ddx}{\frac{d}{dx}} \newcommand{\ddt}{\frac{d}{dt}} \newcommand{\dydx}{\ddxof y} \newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}} \newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}} \newcommand{\dist}{\mathrm{distance}} \newcommand{\arccot}{\mathrm{arccot\:}} \newcommand{\arccsc}{\mathrm{arccsc\:}} \newcommand{\arcsec}{\mathrm{arcsec\:}} \newcommand{\arctanh}{\mathrm{arctanh\:}} \newcommand{\arcsinh}{\mathrm{arcsinh\:}} \newcommand{\arccosh}{\mathrm{arccosh\:}} \newcommand{\sech}{\mathrm{sech\:}} \newcommand{\csch}{\mathrm{csch\:}} \newcommand{\conj}[1]{\overline{#1}} \newcommand{\mathds}{\mathbb} [/math]

• Show directly that [math]\frac{2x-3}{(x-2)^2}[/math] can be written in the form [math]\frac{A}{x-2} + \frac{B}{(x-2)^2}[/math] by first writing [math]\frac{2x-3}{(x-2)^2} = \frac{2(x-2)+1}{(x-2)^2}[/math].
• Following the method in \ref{ex7.4.5a}, show that [math]\frac{ax+b}{(x-k)^2}[/math] can always be written [math]\frac{A}{x-k} + \frac{B}{(x-k)^2}[/math], where [math]A[/math] and [math]B[/math] are constants.
• Extend the result in (b) by factoring, completing the square, and dividing to show directly that
  [[math]] \frac{ax^2+bx+c}{(x-k)^3} \: \mbox{can be written} \: \frac{A}{x-k}+\frac{B}{(x-k)^2}+\frac{C}{(x-k)^3} [[/math]]
  where [math]A[/math], [math]B[/math] and [math]C[/math] are constants.