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Split \\[\\simplify{(({a1+a3})x^2+{a1*a+a1*b+a2+2*a*a3} * x + {a1*a*b + a2*b + a3*a^2})/ ((x + {a})^2 * (x + {b}))}\\] into partial fractions.

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Input the partial fractions here: [[0]].

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Input as the sum of partial fractions.

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Split \\[\\simplify{(({b1+b3})x^2+{b1*p+b1*q+b2+2*p*b3} * x + {b1*p*q + b2*q + b3*p^2})/ ((x + {p})^2 * (x + {q}))}\\] into partial fractions.

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Input the partial fractions here: [[0]].

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Input as the sum of partial fractions.

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Split \\[\\simplify{(({c1+c3})x^2+{c1*p1+c1*q1+c2+2*p1*c3} * x + {c1*p1*q1 + c2*q1 + c3*p1^2})/ ((x + {p1})^2 * (x + {q1}))}\\] into partial fractions.

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Input the partial fractions here: [[0]].

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Input as the sum of partial fractions.

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a)

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We use partial fractions to find $A$, $B$ and $C$ such that: 
$\\simplify{({a1+a3}x^2+{a1*a+a1*b+a2+2*a*a3} * x + {a1*a*b + a2*b + a3*a^2})/ ((x + {a})^2 * (x + {b}))} \\;\\;\\;=\\simplify{A/(x+{a})+B/(x+{a})^2+C/(x+{b})}$

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Dividing both sides of the equation by $\\displaystyle \\simplify[std]{1/( (x+{a})^2(x+{b}) )}\\;\\;$ we obtain:

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$ \\simplify{A(x+{a})(x+{b})+B(x+{b})+C(x+{a})^2 = {a1+a3}*x^2+{a1*a+a1*b+a2+2*a*a3}*x + {a1*a*b + a2*b + a3*a^2}}$

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$\\Rightarrow \\simplify[std]{(A+C)x^2+({a+b}A+B+{2a}C)x+({a*b}A+{b}B+{a*a}C)={a1+a3}*x^2+{a1*a+a1*b+a2+2*a*a3}*x + {a1*a*b + a2*b + a3*a^2}}$

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Identifying coefficients:

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Coefficient $x^2$: $\\simplify[std]{A+C={a1+a3} }$

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Coefficent $x$: $ \\simplify[std]{ {a+b}A+B+{2a}C = {a1*a+a1*b+a2+2*a*a3} }$ 

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Constant term: $\\simplify{{a*b}A+{b}B+{a*a}C ={a1*a*b + a2*b + a3*a^2}}$

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On solving these equations we obtain $A = \\var{a1}$, $B=\\var{a2}$ and $C=\\var{a3}$

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Which gives:$\\simplify{({a1+a3}x^2+{a1*a+a1*b+a2+2*a*a3} * x + {a1*a*b + a2*b + a3*a^2})/ ((x + {a})^2 * (x + {b}))} \\;\\;\\;=\\simplify{{a1}/(x+{a})+{a2}/(x+{a})^2+{a3}/(x+{b})}$

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Apply same method to solve b) and c)

", "contributors": [{"name": "Michael Proudman", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/269/"}]}]}], "contributors": [{"name": "Michael Proudman", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/269/"}]}