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- Jan 3rd 2008, 08:41 AMr_mathsPartial Fractions
- Jan 3rd 2008, 08:48 AMcolby2152
I believe you forgot (to list in your final answer) the first term of one of the original lines in the problem.

- Jan 3rd 2008, 08:50 AMr_maths
My answer is the part in red (Part 3).

- Jan 3rd 2008, 08:58 AMcolby2152
- Jan 3rd 2008, 09:05 AMr_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 - Jan 3rd 2008, 09:18 AMcolby2152
- Jan 3rd 2008, 09:36 AMSoroban
Hello, r_maths!

Quote:

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

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

**not**the original fraction.

But your answer checks out!

$\displaystyle \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}} $

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

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

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

- Jan 3rd 2008, 10:54 AMKrizalid
$\displaystyle \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

$\displaystyle 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 $\displaystyle \frac{{x^3 + 3x^2 + 5x + 4}}

{{(x + 1)(x + 2)}} = x + \frac{1}

{{x + 1}} + \frac{2}

{{x + 2}}$.