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Asks students to find the partil fraction decomposition for a rational function Denominator is a quadratic with distinct factors.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "advice": "We first note that
\\[\\frac{\\simplify{{A+B}x-{A*D+B*C}}}{\\simplify{x^2-{C+D}x+{C*D}}}=\\frac{\\simplify{{A+B}x-{A*D+B*C}}}{(\\simplify{x-{C}})(\\simplify{x-{D}})}
\\]
hence we want to express the function in the form
\\begin{align*}\\frac{\\simplify{{A+B}x-{A*D+B*C}}}{(\\simplify{x-{C}})(\\simplify{x-{D}})}&=\\frac{A}{(\\simplify{x-{C}})}+\\frac{B}{(\\simplify{x-{D}})}\\\\&=\\frac{A(\\simplify{x-{D}})+B(\\simplify{x-{C}})}{(\\simplify{x-{C}})(\\simplify{x-{D}})}\\end{align*}
Equating the numerators we have
\\[\\simplify{{A+B}x-{A*D+B*C}}=A(\\simplify{x-{D}})+B(\\simplify{x-{C}})
\\]
Now letting $x=\\var{D}$ we obtain
\\begin{align*}\\simplify[zeroTerm,!cancelTerms]{{(A+B)*D}-{A*D+B*C}}&=A\\times 0+B\\times\\simplify{{D-C}}\\\\\\simplify{{(A+B)*D}-{A*D+B*C}}&=\\var{D-C}B\\\\B&=\\frac{\\simplify{{(A+B)*D}-{A*D+B*C}}}{\\simplify{{D-C}}}\\\\&=\\var{B}\\end{align*}
Similarly, setting $x=\\var{C}$ we obtain
\\begin{align*}\\simplify[zeroTerm,!cancelTerms]{{(A+B)*C}-{A*D+B*C}}&=A\\times \\simplify{{C-D}}+B\\times 0\\\\\\simplify{{(A+B)*C}-{A*D+B*C}}&=- \\simplify{{C-D}}A\\\\A&=\\frac{{\\simplify{{(A+B)*C}-{A*D+B*C}}}}{\\var{C-D}}\\\\&=\\var{A}\\end{align*}
Hence our partial fraction decomposition is
\\[\\frac{\\simplify{{A+B}x-{A*D+B*C}}}{\\simplify{x^2-{C+D}x+{C*D}}}=\\frac{\\var{A}}{\\simplify{x-{C}}}+\\frac{\\var{B}}{\\simplify{x-{D}}}
\\]
Express \\(\\displaystyle{\\frac{\\simplify{{A+B}x-{A*D+B*C}}}{\\simplify{x^2-{C+D}x+{C*D}}}}\\) in partial fraction form. Show full working on your handwritten working.
\n\\(\\displaystyle{\\frac{\\simplify{{A+B}x-{A*D+B*C}}}{\\simplify{x^2-{C+D}x+{C*D}}}}=\\)[[0]]\\(+\\)[[1]]
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