I am trying to express this as partial fraction:
$\frac{1}{(x+1)(x^2+2x+2)}$
I have a similar exaple that has $5x$ as numerator, it is easy to understand. I do not know what to do with 1 in the numerator, how to solve it?!
I am trying to express this as partial fraction:
$\frac{1}{(x+1)(x^2+2x+2)}$
I have a similar exaple that has $5x$ as numerator, it is easy to understand. I do not know what to do with 1 in the numerator, how to solve it?!
$ \frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2}. $ Then you need to find $A$, $B$, and $C$.
The polynomial $x^2+2x+2$ factors as $(x+1+i)(x+1-i)$, where $i$ is a square root of $-1$. You could go on to write $ \frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2} = \frac{A}{x+1} + \frac{D}{x+1+i} + \frac{E}{x+1-i}, $ and the numbers $D$ and $E$ might not be real.
Later edit in response to a comment: $ \frac{1}{(x+1)(x^2+2x+1)} = \frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2}. $ There are several methods for finding $A$, $B$, and $C$.
Either of two methods begins by multiplying both sides by the denominator, getting $ 1 = A(x^2+2x+2) + (Bx+C)(x+1). $
In one method, you can make the second term vanish by setting $x=-1$: $ 1 = A\Big((-1)^2+2(-1)+2\Big) + \Big(\text{the second term, which is now }0\Big). $ Solving this for $A$ gives $A=1$, and then you write $ 1 = 1(x^2+2x+2) + (Bx+C)(x+1). $ Now you can let $x=1$ in order to make the term involving $B$ vanish: $ 1 = 1(0^2+2\cdot0 + 2) + C(0+1). $ This gives $C=-1$. Then we have $ 1 = (x^2+2x+2) + (Bx-1)(x+1). $ We can no longer make anything vanish without making $B$ disappear, so let x=\text{some number that won't make the arithmetic too messy}. If we let $x=1$, we get $ 1 = (1^2+2\cdot1+2) + (B2-1)(1+1). $ This gives us $B=-1/2$.
Another method is this: We had $ 1 = A(x^2+2x+2) + (Bx+C)(x+1). $ Now multiply this out and collect like terms, where "like" means they are coefficients of the same powers of $x$: $ 0x^2+0x^1 = (A+B)x^2 + (2A+B+C)x + (2A+C). $ Then equate coefficients of common powers of $x$: $ \begin{align} A+B & = 0 \\ 2A+B+C & = 0 \\ 2A+C & = 1 \end{align} $ Then solve that system for $A$, $B$, and $C$.
You set it up in the usual way as $\begin{align*}\frac1{(x+1)(x^2+2x+2)}&=\frac{A}{x+1}+\frac{Bx+C}{x^2+2x+2}\\ &=\frac{A(x^2+2x+2)+(Bx+C)(x+1)}{(x+1)(x^2+2x+2)}\;, \end{align*}$
so that $A(x^2+2x+2)+(Bx+C)(x+1)=1\;.$
Now multiply out the lefthand side to get $(A+B)x^2+(2A+B+C)x+(2A+C)=1$
and equate coefficients of $x^2,x$, and $1$ to get (in that order):
$\left\{\begin{align*} &A+B=0\\ &2A+B+C=0\\ &2A+C=1 \end{align*}\right.$
Then solve the system for $A,B$, and $C$.
You want to find $A, B,$ and $C$ such that
$\frac{1}{(x+1)(x^2 + 2x + 2)} = \frac{A}{x+1} + \frac{Bx + C}{x^2 + 2x + 2} $ That is such that $\begin{align}0x^2 + 0x + 1 &= A(x^2 + 2x + 2) + (x+1)(Bx+c)\\ &= (A+B)x^2 + (2A+B+C)x + 2A + C. \end{align}$ So you get three equations $\begin{align} 0 &= A + B \\ 0 &= 2A + B + C \\ 1 &= 2A + C. \end{align}$ Solving this I get $A =1, B = -1, C= -1$.