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Rewrite the expression $\\frac{{m}x^2+{n}x+{p}}{x+a}$ as partial fractions in the form $\\frac{A}{x+a}+Bx+C$.
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Rewrite the following expression as partial fractions:
\n\\[ \\simplify{({m}x^2+{n}x+{p})/(x+{a})} .\\]
", "advice": "When calculating the partial fractions for an improper fraction (the numerator has an equal or higher degree than the denominator), an extra polynomial is added to the partial fractions that would normally be calculated. The added polynomial has degree $n-d$, where $n$ is the degree of the numerator and $d$ is the degree of the denominator.
\nSince \\[ \\simplify{({m}x^2+{n}x+{p})/(x+{a})} \\] has a numerator with degree 2 and a denominator with degree 1, we need to add a polynomial with degree 1. Therefore,
\n\\[\\simplify{({m}x^2+{n}x+{p})/(x+{a})} = \\simplify{A/(x+{a}) + B*x+C} .\\]
\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{{m}x^2+{n}x+{p} = A + B*x(x+{a}) + C(x+{a})}.\\]
\n\nTo find $A$, we can eliminate $B$ and $C$ by setting $x=\\var{-a}$:
\n\\[ \\simplify{{m*a^2}-{n*a}+{p} = A}.\\]
\nTo find $B$ and $C$, we can compare the coefficients of the $x^2$-terms and $x$-terms:
\n\\[ \\begin{split} (x^2)&: \\quad \\var{m} &\\,= B \\\\ (x)&: \\quad \\var{n} &\\,= \\simplify{{a}B+C}. \\end{split} \\]
\nHence,
\n\\[ B=\\var{m} \\, ,\\, C=\\simplify{{n-m*a}}, \\]
\nand
\n\\[ \\simplify{({m}x^2+{n}x+{p})/(x+{a})} = \\simplify{{m*a^2-n*a+p}/(x+{a}) + {m}x + {n-m*a}}. \\]
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