PARTIAL FRACTIONS- HELP!
- Thread starter KingAce
- Start date
\(\displaystyle \L\\\frac{A}{x-3}+\frac{B}{x-2}=x-8\)
\(\displaystyle \L\\A(x-2)+B(x-3)=x-8\)
\(\displaystyle \L\\Ax-2A+Bx-3B=x-8\)
Equate coefficients:
\(\displaystyle \L\\-2A-3B=-8, \;\ A+B=1\)
Solving, we find:
\(\displaystyle \L\\A=-5 \;\ and \;\ B=6\)
\(\displaystyle \H\\\frac{-5}{x-3}+\frac{6}{x-2}\)
\(\displaystyle \L\\A(x-2)+B(x-3)=x-8\)
\(\displaystyle \L\\Ax-2A+Bx-3B=x-8\)
Equate coefficients:
\(\displaystyle \L\\-2A-3B=-8, \;\ A+B=1\)
Solving, we find:
\(\displaystyle \L\\A=-5 \;\ and \;\ B=6\)
\(\displaystyle \H\\\frac{-5}{x-3}+\frac{6}{x-2}\)
Hello, KingAce!
If no one has taught you how to do this, I'll baby-step through it for you
. . and show you an alternate method.
You already know that the denominator factors: \(\displaystyle \,x^2\,-\,5x\,+\,6\:=\
x\,-\,2)(x\,-\,3)\)
We assume we can break it up into two fractions like this:
. . \(\displaystyle \L\frac{x\,-\,8}{(x\,-\,2)(x\,-\,3)} \;=\;\frac{A}{x\,-\,2}\,+\,\frac{B}{x\,-\,3}\)
Multiply through by the LCD: \(\displaystyle \L\:x\,-\,8\;=\;A(x\,-\,3)\,+\,B(x\,-\,2)\)
. . and we must determine \(\displaystyle A\) and \(\displaystyle B.\)
We have only one equation ... How can we solve for two unknowns?
Simple ... plug in various values for \(\displaystyle x\)
. . and we can create as many equations as we want.
\(\displaystyle \begin{array}{cccc}\text{Let }x\,=\,2: \;\; 2\,-\,8\:=\:A(2\,-\,3)\,+\,B(2\,-\,2) & \;\Rightarrow\; &-6\:=\:A(-1)\,+\,B(0) & \;\Rightarrow\; &A\,=\,6 \\ \\ \\
\text{Let }x\,=\,3: \;\; 3\,-\,8\:=\:A(3\,-\,3)\,+\,B(3\,-\,2) & \;\Rightarrow\; & -5\:=\:A(0)\,+\,B(1) & \;\Rightarrow\; & B\,=\,-5 \end{array}\)
Therefore: \(\displaystyle \L\:\frac{x\,-\,8}{(x\,-\,2)(x\,-\,3)}\;=\;\frac{6}{x\,-\,2} \,-\,\frac{5}{x\,-\,3}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
You can check this result yourself: subtract the two fractions.
. . (It's good practice.)
If no one has taught you how to do this, I'll baby-step through it for you
. . and show you an alternate method.
You already know that the denominator factors: \(\displaystyle \,x^2\,-\,5x\,+\,6\:=\
x\,-\,2)(x\,-\,3)\)We assume we can break it up into two fractions like this:
. . \(\displaystyle \L\frac{x\,-\,8}{(x\,-\,2)(x\,-\,3)} \;=\;\frac{A}{x\,-\,2}\,+\,\frac{B}{x\,-\,3}\)
Multiply through by the LCD: \(\displaystyle \L\:x\,-\,8\;=\;A(x\,-\,3)\,+\,B(x\,-\,2)\)
. . and we must determine \(\displaystyle A\) and \(\displaystyle B.\)
We have only one equation ... How can we solve for two unknowns?
Simple ... plug in various values for \(\displaystyle x\)
. . and we can create as many equations as we want.
\(\displaystyle \begin{array}{cccc}\text{Let }x\,=\,2: \;\; 2\,-\,8\:=\:A(2\,-\,3)\,+\,B(2\,-\,2) & \;\Rightarrow\; &-6\:=\:A(-1)\,+\,B(0) & \;\Rightarrow\; &A\,=\,6 \\ \\ \\
\text{Let }x\,=\,3: \;\; 3\,-\,8\:=\:A(3\,-\,3)\,+\,B(3\,-\,2) & \;\Rightarrow\; & -5\:=\:A(0)\,+\,B(1) & \;\Rightarrow\; & B\,=\,-5 \end{array}\)
Therefore: \(\displaystyle \L\:\frac{x\,-\,8}{(x\,-\,2)(x\,-\,3)}\;=\;\frac{6}{x\,-\,2} \,-\,\frac{5}{x\,-\,3}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
You can check this result yourself: subtract the two fractions.
. . (It's good practice.)
Count Iblis
Junior Member
- Joined
- Feb 13, 2007
- Messages
- 108
A faster way to obtain the answer is as follows. You know that:
\(\displaystyle \L \frac{x-8}{x^2 - 5x + 6} = \frac{A}{x-2} + \frac{B}{x-3}\)
Multiply both sides by \(\displaystyle x-2\) and take the limit \(\displaystyle x\rightarrow 2\) to obtain \(\displaystyle A\). To obtain \(\displaystyle B\) you multiply by \(\displaystyle x-3\) and take the limit \(\displaystyle x\rightarrow 3\). The limits are easily evaluated using L'Hopital's rule.
\(\displaystyle \L \frac{x-8}{x^2 - 5x + 6} = \frac{A}{x-2} + \frac{B}{x-3}\)
Multiply both sides by \(\displaystyle x-2\) and take the limit \(\displaystyle x\rightarrow 2\) to obtain \(\displaystyle A\). To obtain \(\displaystyle B\) you multiply by \(\displaystyle x-3\) and take the limit \(\displaystyle x\rightarrow 3\). The limits are easily evaluated using L'Hopital's rule.
Count Iblis
Junior Member
- Joined
- Feb 13, 2007
- Messages
- 108
Actually, it is even simpler than that and there is no need to use L'Hopital's rule. Since you already have the factorization of the denominator, the limits are trivial. If you e.g. multiply by \(\displaystyle x-2\) you are left with the factor \(\displaystyle x-3\) in the denominator.