Please check my Partial Fraction solution.

[nil101]

Please can you check where I have gone wrong with this solution. The given answer is different from mine.

Resolve into Partial Fractions

\(\displaystyle \L\frac{{2x}}{{(1 + 2x)(1 + 3x^2 )}}\)


\(\displaystyle \L
\frac{{2x}}{{(1 + 2x)(1 + 3x^2 )}} \equiv \frac{A}{{1 + 2x}} + \frac{{Bx + C}}{{1 + 3x^2 }}

\\
\begin{array}{l}
\\
\equiv \frac{{A(1 + 3x^2 ) + (Bx + C)(1 + 2x)}}{{(1 + 2x)(1 + 3x^2 )}} \\
\\
\end{array}
\\

\begin{array}{l}
{\rm multiplying out numerator and grouping coefficients }... \\
\\
\end{array}
\\
\\
\begin{array}{l}
\\
\equiv {\rm (3A + 2B)}x^2 + (B + 2C)x + (A + C) \\

\\
{\rm comparing coefficients:} \\
\\
{\rm [}x^2 {\rm ] }0 = 3A + 2B{\rm }.............\left( {\rm 1} \right) \\
{\rm [}x{\rm ] }2 = B + 2C{\rm }.............\left( {\rm 2} \right) \\
{\rm[} CT{\rm ] } 0 = A + C{\rm }...............\left( {\rm 3} \right) \\

\left( {\rm 2} \right){\rm x }2{\rm }4 = 2B + 4C \\
{\rm minus}\left( {\rm 1} \right){\rm }\underline {{\rm 0 = 3A + 2B}} \\
{\rm 4 = 3A + 4C }................\left( {\rm 4} \right) \\
\\
\left( {\rm 3} \right){\rm x 3 }\underline {0 = 3A + 3C} {\rm }.....{\rm take from }\left( {\rm 4} \right) \\
{\rm }4 = C \\
\\
{\rm substituting into }\left( {\rm 3} \right){\rm and }\left( {\rm 2} \right),{\rm A = } - 4{\rm and }B = - 6 \\
\end{array}

\\
\\
\equiv \frac{{2x}}{{(1 + 2x)(1 + 3x^2 )}} \equiv \frac{{ - 4}}{{(1 + 2x)}} - \frac{{6x + 4}}{{(1 + 3x^2 )}}
\\
\\\)




But the given answer is \(\displaystyle \L \frac{{2x}}{{(1 + 2x)(1 + 3x^2 )}} \equiv \frac{{ - 4}}{{7(1 + 2x)}} - \frac{{6x + 4}}{{7(1 + 3x^2 )}}\)

What have I missed?

Thanks

 

[galactus]

\(\displaystyle \L\\A(1+3x^{2})+(1=2x)(Bx+C)\rightarrow\)

\(\displaystyle \L\\A+3Ax^{2}+Bx+C+2Bx^{2}+2Cx=2x\)

Equating coefficents leads to:

\(\displaystyle A+C=0.......[1]\\3A+2B=0.......[2]\\B+2C=2.......[3]\)

Solve [3] for B:

B=2-2C

Sub into [2]:

3A+2(2-2C)=0

3A+4-4C=0

A=-C

3(-C)+4-4C=0

C=4/7
B=6/7
A=-4/7

\(\displaystyle \L\\\frac{-4}{7(1+2x)}+\frac{4}{7(3x^{2}+1)}+\frac{6x}{7(3x^{2}+1})\)

=\(\displaystyle \L\\\frac{-4}{7(2x+1)}+\frac{6x+4}{7(3x^{2}+1)}\)

 

[soroban]

Hello, nil101!

There is another (easier) way to solve Partial Fractions.
Many teachers don't teach it . . . or don't know it!
I'll demonstrate with a simpler example.

Resolve \(\displaystyle \L\,\frac{8x}{(x\,-\,1)(x\,+\,3)\) in partial fractions.

We start off the same way: \(\displaystyle \L\:\frac{8x}{(x\,-\,1)(x\,+\,3)}\:=\:\frac{A}{x\,-\,1}\,+\,\frac{B}{x\,+\,3}\)

Multiply by the LCD: \(\displaystyle \L\:8x\;=\;A(x\,+\,3)\,+\,B(x\,-\,1)\)


Now we substitute "good" values of \(\displaystyle x\) *

Let \(\displaystyle x\,=\,1:\;\;8(1)\;=\;A(1\,+\,3)\,+\,B(0)\)
\(\displaystyle \;\;\)and we have: \(\displaystyle \,8\,=\,4A\;\;\Rightarrow\;\;A\,=\,2\)

Let \(\displaystyle x\,=\,-3:\;\;8(-3)\;=\;A(0)\,+\,B(-3\,-\,1)\)
\(\displaystyle \;\;\)and we have: \(\displaystyle \,-24\,=\,-4B\;\;\Rightarrow\;\;B\,=\,6\)

And we have: \(\displaystyle \L\,\frac{8x}{(x\,-\,1)(x\,+\,3)}\;=\;\frac{2}{x\,-\,1}\,+\,\frac{6}{x\,+\,3}\)

*I assume you caught on what a "good" value is.
\(\displaystyle \;\;\;\)It's a value of \(\displaystyle x\) that "zeros out" one or more variables.

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Now let's try your problem . . .

Resolve into Partial Fractions: \(\displaystyle \L\:\frac{2x}{(1 + 2x)(1 + 3x^2 )}\)

We have: \(\displaystyle \L\:\frac{2x}{(1\,+\,2x)(1\,+\,3x^2)}\;=\;\frac{A}{1\,+\,2x}\,+\,\frac{Bx}{1\,+\,3x^2}\,+\,\frac{C}{1\,+\,3x^2}\)

\(\displaystyle \;\;\)Note that I immediately make two fractions for the quadratic factor.

Multiply by LCD: \(\displaystyle \L\:2x\;=\;A(1\,+\,3x^2)\,+\,Bx(1\,+\,2x)\,+\,C(1\,+\,2x)\)


Substitute some "good" values of \(\displaystyle x\) . . . well, we do the best we can. **

Let \(\displaystyle x\,=\,-\frac{1}{2}:\;\;2\left(-\frac{1}{2}\right)\;=\;A\left(1\,+\,3\left[-\frac{1}{2\right]^2\right) \,+\,B\left(-\frac{1}{2}\right)(0)\,+\,C(0)\;\;\Rightarrow\;\;\frac{7}{4}A\,=\,-1\;\;\Rightarrow\;\;A\,=\,-\frac{4}{7}\)

Let \(\displaystyle x\,=\,0:\;\;2(0)\:=\:A(1+0)\,+\,B(0)\,+\,C(1+0)\;\;\Rightarrow\;\;A\,+\,C\:=\:0\)
\(\displaystyle \;\;\)Since \(\displaystyle A\,=\,-\frac{4}{7}\), we have: \(\displaystyle \,-\frac{4}{7}\,+\,C\:=\:0\;\;\Rightarrow\;\;C\,=\,\frac{4}{7}\)

Let \(\displaystyle x\,=\,1:\;\;2(1)\;=\;a(1+3)\,+\,B(1)(3)\,+\,C(3)\;\;\Rightarrow\;\;4A\,+\,3B\,+\,3C\:=\:2\)
\(\displaystyle \,\,\)Since \(\displaystyle A\,=\,-\frac{4}{7},\;C\,=\,\frac{4}{7}\), we have: \(\displaystyle \,4\left(-\frac{4}{7}\right)\,+\,3B\,+\,3\left(\frac{4}{7}\right)\:=\:2\;\;\Rightarrow\;\;B\,=\,\frac{6}{7}\)

Therefore: \(\displaystyle \L\:\frac{2x}{(1\,+\,2x)(1\,+\,3x^2)}\;=\;\frac{-\frac{4}{7}}{1\,+\,2x}\,+\,\frac{\frac{6}{7}x}{1\,+\,3x^2}\,+\,\frac{\frac{4}{7}}{1\,+\,3x^2}\)

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**
If all the factors are linear and distinct, there are always enough "good" values of \(\displaystyle x\).

If there are repated linear factors or a quadratic factor,
\(\displaystyle \;\;\)there aren't enough "good" values.
So we pick any convenient values of \(\displaystyle x\) and solve anyway.
\(\displaystyle \;\;\)We usually have to substitute-in one of the known values.