# Partial Fractions

• January 3rd 2008, 08:41 AM
r_maths
Partial Fractions
• January 3rd 2008, 08:48 AM
colby2152
I believe you forgot (to list in your final answer) the first term of one of the original lines in the problem.
• January 3rd 2008, 08:50 AM
r_maths
My answer is the part in red (Part 3).
• January 3rd 2008, 08:58 AM
colby2152
Quote:

Originally Posted by r_maths
My answer is the part in red (Part 3).

Hmm, I get $A = 1, B = 2$, which leads to a final answer of: $x+\frac{1}{x+1}+\frac{2}{x+2}$. I do not see how the x term is not supposed to be there.
• January 3rd 2008, 09:05 AM
r_maths
I did include the x :confused:

I got the same answer as you, but the answer from the book gives a different answer...

http://img526.imageshack.us/img526/3820/33966436ey9.png
• January 3rd 2008, 09:18 AM
colby2152
Quote:

Originally Posted by r_maths
I did include the x :confused:

I got the same answer as you, but the answer from the book gives a different answer...

http://img526.imageshack.us/img526/3820/33966436ey9.png

I do not see how B equals one, so the book is probably wrong unless both of us missed something.
• January 3rd 2008, 09:36 AM
Soroban
Hello, r_maths!

Quote:

Express $\frac{x^3+3x^2+5x+4}{(x+1)(x+2)}$ in partial fractions.

$\frac{x^3+3x^2+5x+4}{(x+1)(x+2)} \;=\;x + \frac{3x+4}{(x+1)(x+2)} \;=\;x + \frac{1}{x+1} + \frac{2}{x+2}\quad{\color{red}Right!}$

Book answer: . $x + \frac{1}{x+1} + \frac{1}{x+2}\quad{\color{red}Wrong!}$

$\text{Their answer adds up to: }\;\frac{x^3+3x^2+4x+3}{(x+1)(x+2)}$ . . . . obviously not the original fraction.

$\frac{x}{1}\cdot{\color{blue}\frac{(x+1)(x+2)}{(x+ 1)(x+2)}} + \frac{1}{x+1}\cdot{\color{blue}\frac{x+2}{x+2}} + \frac{2}{x+2}\cdot{\color{blue}\frac{x+1}{x+1}}$

. . $= \;\frac{x(x+1)(x+2) + (x+2) + 2(x+1)}{(x+1)(x+2)}$

. . $= \;\frac{x^3+3x^2+2x + x + 2 + 2x + 2}{(x+1)(x+2)}$

. . $= \;\frac{x^3+3x^2+5x+4}{(x+1)(x+2)}$ . . . . See?

• January 3rd 2008, 10:54 AM
Krizalid
$\frac{{x^3 + 3x^2 + 5x + 4}}
{{(x + 1)(x + 2)}} = \frac{{x^3 + 3x^2 + 2x + 3x + 4}}
{{(x + 1)(x + 2)}} = \frac{{x(x + 1)(x + 2) + (3x + 4)}}
{{(x + 1)(x + 2)}}$
.

Now

$x + \frac{{3x + 4}}
{{(x + 1)(x + 2)}} = x + \frac{{(x + 2) + 2(x + 1)}}
{{(x + 1)(x + 2)}} = x + \frac{1}
{{x + 1}} + \frac{2}
{{x + 2}}$
.

Hence $\frac{{x^3 + 3x^2 + 5x + 4}}
{{(x + 1)(x + 2)}} = x + \frac{1}
{{x + 1}} + \frac{2}
{{x + 2}}$
.