1. ## Partial Fractions

2. I believe you forgot (to list in your final answer) the first term of one of the original lines in the problem.

3. My answer is the part in red (Part 3).

4. 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.

5. I did include the x

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

6. Originally Posted by r_maths
I did include the x

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

I do not see how B equals one, so the book is probably wrong unless both of us missed something.

7. Hello, r_maths!

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?

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

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