Title: Partial Fractions
Course: Demonstrating STACK (v4.4.7)

Express \[[value],\] as a sum of partial fractions.
Feedback: Below are the rules of partial fractions.[[facts:alg_partial_fractions]] We note that the numerator \[[value]\] is cubic and that the denominator \[[value]=[value]\] is quadratic. Dividing [value] by [value] we obtain a linear term of the form [value] and a linear remainder of the form [value] and \[[value] = ([value])([value])+[value].\] Expanding out the brackets we find \[[value] = [value]+[value].\] Matching coefficients we see that \(g=[value]\) and that \(h=[value]-[value]=[value]-[value]=[value]\). Also \[\begin{array}{rcl}\displaystyle h+m+[value]=[value] + m + [value]&\displaystyle =&\displaystyle [value]\mbox{,}\\ [value]+n=[value]+n&\displaystyle =&\displaystyle [value]\mbox{.} \end{array}\] Therefore \(m=[value]\) and \(n=[value]\). The quotient is \([value]=[value]\) and the remainder is \([value]=[value]\). As a sum of partial fractions \[ [value]=\frac{A}[value]+\frac{B}[value]\mbox{.}\] Multiplying both sides through by [value] we find: \[[value]= [value] + B\mbox{.}\] Matching coefficients we see that \(A=[value]\) and setting \(x=[value]\) we find that \(B=[value]\) so \[ [value]=\frac[value][value]+\frac[value][value]\mbox{.}\] Therefore \[[value]=[value]\mbox{.}\]

Express \[[value]\] as a sum of a linear function and a rational function with a linear numerator and denominator [value].
Feedback: Below are the rules of partial fractions.[[facts:alg_partial_fractions]] We note that the numerator [value] is cubic and that the denominator [value] is quadratic. Dividing [value] by [value] we obtain a quotient with a linear numerator of the form [value] and a linear term of the form [value], that is \[[value]=\frac[value][value]+[value].\] Multiplying through by [value] \[[value] = ([value])([value])+[value].\] Expanding out the brackets we find \[[value] = [value]+[value].\] Matching coefficients in \(x^3\) we see \(m=[value]\). Matching coefficients in lower order terms and substituting we find that \(n=[value]\), \(g=[value]\) and \(h=[value]\). Therefore \[[value]=[value].\]

Express

\[[value]\]

as a sum of partial fractions.
Feedback: Below are the rules of partial fractions.[[facts:alg_partial_fractions]] As a sum of partial fractions \[ [value]=\frac{A}{x}+\frac{B}[value]+\frac{C}[value],\] and multiplying both sides through by [value] we find \[\begin{array}{rcl}\displaystyle [value]&\displaystyle =&\displaystyle [value] + [value] + [value],\\ &\displaystyle =&\displaystyle [value] + [value] + [value] + [value] + [value] + [value],\\ &\displaystyle =&\displaystyle [value] + [value] + [value]. \end{array}\] Matching coefficients \[\begin{array}{rcl}\displaystyle A + B&\displaystyle =&\displaystyle [value]\,\\ [value] + C&\displaystyle =&\displaystyle 0,\\ [value]&\displaystyle =&\displaystyle [value]. \end{array}\] So, \[\begin{array}{rcl}\displaystyle A&\displaystyle =&\displaystyle [value],\\ B&\displaystyle =&\displaystyle [value] - A = [value] - [value],\\ &\displaystyle =&\displaystyle [value],\\ C&\displaystyle =&\displaystyle [value] = [value] - [value],\\ &\displaystyle =&\displaystyle [value]. \end{array}\] Therefore, \[[value]=[value].\]

Express \[[value]\] as a sum of partial fractions.
Feedback: Below are the rules of partial fractions.[[facts:alg_partial_fractions]] So, given \[ [value]=\frac{A}[value] + \frac{B}[value] + \frac{C}[value],\] then multiplying through by the denominator [value] we have that \[ [value] = [value] + [value]+[value]. \] Setting \(x = [value]\) we see that \(A = [value]\) and letting \(x = [value]\) we find that \(C = [value]\). Putting \(x = 0\) we can also see that \(B = [value]\). Therefore \[ [value]=[value].\]

Give the expression \[[value]\] as a sum of partial fractions.
Feedback: Below are the rules of partial fractions.[[facts:alg_partial_fractions]] Applying these rules, we find that \[ [value] = \frac{A}[value] + \frac{B}[value],\] and multiplying through by the denominator [value] we see that: \[ [value] = [value] + [value]. \] Setting \(x = [value]\) we see that \(A = [value]\) and setting \(x = [value]\) we see that \(B = [value]\). Therefore, \[ [value] = [value].\]

Express

\[[value]\mbox{,}\]

as a sum of partial fractions.
Feedback: Below are the rules of partial fractions.[[facts:alg_partial_fractions]] Applying these rules, we find that: \[ [value]=\frac{A}{x}+\frac{B}[value]+\frac{C}[value]\mbox{.}\] Multiplying both sides through by [value] \[\begin{array}{rcl}\displaystyle [value]&\displaystyle =&\displaystyle [value] + [value] + [value]\mbox{,}\\ [value]&\displaystyle =&\displaystyle [value] + [value] + [value]\mbox{.} \end{array}\] Matching coefficients \[\begin{array}{rcl}\displaystyle [value]&\displaystyle =&\displaystyle 0\mbox{,}\\ [value]&\displaystyle =&\displaystyle 0\mbox{,}\\ [value]&\displaystyle =&\displaystyle [value]\mbox{.} \end{array}\] So \[A=[value]\mbox{.}\] Substituting this value in the other two equations we see that \[\begin{array}{rcl}\displaystyle [value]&\displaystyle =&\displaystyle 0\mbox{,}\\ [value]&\displaystyle =&\displaystyle 0\mbox{.} \end{array}\] Rearranging \[\begin{array}{rcl}\displaystyle [value] &\displaystyle =&\displaystyle [value]\mbox{,}\\ [value] &\displaystyle =&\displaystyle [value]\mbox{.} \end{array}\] Multiplying the first of these two equations through by [value] we obtain: \[ [value]=[value]\mbox{,} \] and subtracting the second of the pair of equations above from this equation we can eliminate C to give an equation only in B. We find that \[\begin{array}{rcl}\displaystyle B&\displaystyle =&\displaystyle [value]\mbox{,}\\ C&\displaystyle =&\displaystyle [value]\mbox{.} \end{array}\] Therefore \[[value]=[value].\]

Give the expression

\[[value]\]

as a sum of partial fractions.
Feedback: Below are the rules of partial fractions.[[facts:alg_partial_fractions]] Applying these rules, we find that \[ [value] = \frac{A}{x} + \frac{B}[value] + \frac{C}[value],\] and multiplying through by the denominator [value] we see that \[ [value] = [value] + [value] + [value]. \] Setting \(x = 0\) we see that \(A = [value]\), setting \(x= [value]\) we see that \(B = [value]\) and setting \(x = [value]\) we see that \(C = [value]\). Therefore \[ [value] = [value].\]

Select the appropriate format for expressing the following rational function in partial fractions. \[ f(x) = [value] \] \[ (1) = \frac{A}{x-[value]} + \frac{B}{x-[value]} + \frac{C}[value],\quad (2) = \frac{A}{(x-[value])^2} + \frac{B}{x-[value]} + \frac{C x + D}[value], \] \[ (3) = \frac{A}{(x-[value])^2} + \frac{B}{x-[value]} + \frac{C}[value],\quad (4) = \frac{A}{x-[value]} + \frac{B}{x-[value]} + \frac{C x + D}[value]. \] Format
Calculate the unknown coefficients. Enter these in the correct order as a list, i.e. \([A,B,C,D]\) below, inserting a zero for any unused coefficients.
Feedback: Below are the rules of partial fractions.[[facts:alg_partial_fractions]] \[ f(x) = [value] \] To express as partial fractions let\[ [value]= \frac{A}[value]+\frac{B}[value]+\frac{Cx+D}[value], \mbox{ which is format 2.}\] Multiplying throughout by [value] gives \[[value]=A([value])+B([value])([value])+(Cx+D)[value]\] Let \(x = [value]\), therefore \[[value]=[value]\] \[A=\frac[value][value]=[value]\] Let \(x = 0\), therefore \[[value]=A([value])+B([value])([value])+D([value])=[value]\] \[[value] =[value]\] Equate coefficients of \(x^3\) \[[value] = B+C\] Equate coefficients of \(x^2\) \[[value] = [value]=[value]\] \[[value] = [value]\] Solving the 4 equations in \(A,B,C\) and \(D\) gives \[[value]\]