// Numbas version: finer_feedback_settings {"name": "Partial Fractions: Proper Fractions 5a", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Partial Fractions: Proper Fractions 5a", "tags": [], "metadata": {"description": "

Rewrite the expression $\\frac{n}{(x+a)(x+b)(x+c)}$ as partial fractions in the form $\\frac{A}{x+a}+\\frac{B}{x+b}+\\frac{C}{x+c}$.

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Rewrite the following expression as partial fractions:

\n

\\[ \\simplify{{n}/((x+{a})(x+{b})(x+{c}))}. \\]

", "advice": "

To express \\[ \\simplify{{n}/((x+{a})(x+{b})(x+{c}))} \\] as partial fractions, we want to set this equal to the sum of three fractions with denominators $\\simplify{x+{a}}$, $\\simplify{x+{b}}$, and $\\simplify{x+{c}}$. Since these are all distinct linear factors, this tells us that the numerators will be constants, which we will call $A$, $B$, and $C$:

\n

\\[ \\simplify{{n}/((x+{a})(x+{b})(x+{c}))} = \\simplify{A/(x+{a}) + B/(x+{b})+ C/(x+{c})}.\\]

\n

To find the values of $A$, $B$, and $C$, we want to multiply this equation by the denominator of the left-hand side. This gives

\n

\\[ \\simplify{{n}=A(x+{b})(x+{c})+B(x+{a})(x+{c}) + C(x+{a})(x+{b})}.\\]

\n

There are two methods of finding $A$, $B$, and $C$. The first is to choose suitable values for $x$ which will eliminate two of the terms, and the other is to compare the coefficients of each side of the equation. We will cover both methods here.

\n

Method 1:

\n

To find $A$, we can eliminate $B$ and $C$ by setting $x=\\var{-a}$:

\n

\\[ \\simplify{{n}=A{(b-a)(c-a)}} \\implies A=\\simplify[fractionNumbers]{{Asol}}.\\]

\n

Finding $B$ by setting $x=\\var{-b}$:

\n

\\[ \\simplify{{n}=B{(a-b)(c-b)}} \\implies B=\\simplify[fractionNumbers]{{Bsol}}.\\]

\n

Finally, setting $x=\\var{-c}$ we can find C:

\n

\\[ \\simplify{{n}=C{(a-c)(b-c)}} \\implies C=\\simplify[fractionNumbers]{{Csol}}.\\]

\n

Therefore,

\n

{check}

\n

\n

\n

Method 2:

\n

By comparing the coefficients of the $x^2$-terms, $x$-terms and the constant terms we can form a set of simultaneous equations to find $A$, $B$ and $C$.

\n

\\[ \\begin{split} \\simplify{{n}} &\\,= \\simplify{A(x+{b})(x+{c})+B(x+{a})(x+{c}) + C(x+{a})(x+{b})} \\\\ &\\,= \\simplify{A(x^2+{b+c}x+{b*c})+B(x^2+{a+c}x+{a*c})+C(x^2+{a+b}x+{a*b})} \\\\ &\\,=\\simplify{(A+B+C)x^2+({b+c}A+{a+c}B+{a+b}C)x+{b*c}A+{a*c}B+{a*b}C} . \\end{split} \\]

\n

\\[ \\begin{split}(x^2)&: \\quad 0 &\\,= \\simplify{A+B+C} \\\\ (x)&: \\quad 0 &\\,= \\simplify{{b+c}A+{a+c}B+{a+b}C} \\\\ (c)&:\\quad \\var{n} &\\,= \\simplify{{b*c}A+{a*c}B+{a*b}C} .\\end{split} \\]

\n

Hence,

\n

\\[A=\\simplify[fractionNumbers]{{Asol}},\\,B=\\simplify[fractionNumbers]{{Bsol}}, \\, C=\\simplify[fractionNumbers]{{Csol}}, \\]

\n

and

\n

{check}

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