Rewrite the expression $\\frac{nx+k}{(x+a)(x+b)^2}$ as partial fractions in the form $\\frac{A}{x+a}+\\frac{B}{x+b}+\\frac{C}{(x+b)^2}$.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Rewrite the following expression as partial fractions:
\n\\[ \\simplify{({n}x+{k})/((x+{a})(x+{b})^2)}. \\]
", "advice": "To express \\[ \\simplify{({n}x+{k})/((x+{a})(x+{b})^2)} \\] in terms of partial fractions, we want to set this equal to the sum of three fractions with denominators $\\simplify{x+{a}}$, $\\simplify{x+{b}}$, and $\\simplify{(x+{b})^2}$. Since we have a distinct linear factor and a repeated linear factor, this tells us that the numerators will be constants, which we will call $A$, $B$, and $C$:
\n\\[ \\simplify{({n}x+{k})/((x+{a})(x+{b})^2)} = \\simplify{A/(x+{a}) + B/(x+{b})+ C/(x+{b})^2}.\\]
\nTo find the values of $A$, $B$, and $C$, we want to first multiply this equation by the denominator of the left-hand side. This gives
\n\\[ \\simplify{{n}x+{k}=A(x+{b})^2+B(x+{a})(x+{b}) + C(x+{a})}.\\]
\n(Note: To find $A$, $B$, and $C$, we will use a combination of choosing suitable values of $x$ to eliminate terms, and equating coefficients. It can be solved by only equating coefficients, but this is a more efficient process.)
\n\nTo find $A$, we can eliminate $B$ and $C$ by setting $x=\\var{-a}$:
\n\\[ \\simplify{{k-n*a}=A{(b-a)^2}} \\implies A=\\simplify[fractionNumbers]{{Asol}}.\\]
\nTo find $C$, we can eliminate $A$ and $B$ by setting $x=\\var{-b}$:
\n\\[ \\simplify{{k-n*b}=C{(a-b)}} \\implies C=\\simplify[fractionNumbers]{{Csol}}.\\]
\nFinally, by equating coefficients of the $x^2$-terms we can find $B$:
\n\\[ (x^2): \\quad 0 = \\simplify{A+B} \\implies B=-A. \\]
\nTherefore, \\[ B=\\simplify[fractionNumbers]{{Bsol}}, \\]
\nand
\n{check}
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\\\\[ \\\\simplify{({n}x+{k})/((x+{a})(x+{b})^2)} = \\\\simplify{{Asol}/(x+{a})+{Bsol}/(x+{b})+{Csol}/(x+{b})^2}.\\\\]
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