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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?!

Thomas
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Dumbo
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  • you don't need to deal with numerator expression much just in case if numerator polynomial degree is equal or greater than denominator polynomial degree than you have to convert it into proper fraction(proper fraction is a fraction whose numerator degree is less than its denominator degree) than use conventional approach to solve partial fraction – iostream007 May 10 '13 at 07:18

3 Answers3

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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 $C$.

Brian M. Scott
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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 $A =1, B = -1, C= -1$.

Thomas
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$$ \frac{A}{x+1} + \frac{Bx+C}{x^2+2x+2}. $$ 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 $C$.

Michael Hardy
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  • I knew this :P I want to know what does the equation look like, I mean the equations to find A, B and C. is it 1 = A()B()C() ? – Dumbo Apr 11 '12 at 21:59
  • You might want to check some of my arithmetic....... – Michael Hardy Apr 11 '12 at 22:57
  • Guess you had to put x=0 to vanish B, is that right?! – Dumbo Apr 12 '12 at 21:12
  • @MichaelHardy decomposing a fraction of the form $\frac{1}{p(z)}$ where p(z) is a polynomial on C will it always gives me franctions with Constants as numerators? – Jam Aug 19 '16 at 12:59
  • @ManolisLyviakis : It will if you use complex numbers, but if you are restricted to real numbers you will sometimes get an irreducible quadratic factor in the denominator like the one above, and then the numerator will by of the form $Bx+C$, which is constant only if $B=0$. The polynomial $x^2+2x+2$ is irreducible when one is restricted to real numbers. One sees that by looking at the discriminant $b^2-4ac = 2^2 - 4\cdot1\cdot(-2) = -4$. That number has no square root among the real numbers. But you can factor $x^2+2x+2$ if complex numbers are used. $\qquad$ – Michael Hardy Aug 19 '16 at 18:56
  • @MichaelHardy oh i got some things what i ask more precisely is why when the polynomial can be factor you get constants on the numerator!! – Jam Aug 19 '16 at 19:36
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    @ManolisLyviakis : If you get a non-constant numerator, as in $\dfrac{5x+4}{2x+3}$, then you can do long division: $$ \begin{array}{ccccccc} & & \frac 5 2 \ \ 2x+3 & ) & 5x & + & 4 \ & & 5x & + & \frac{15} 2 \ \ & & & & \frac{-7} 2 \end{array} $$ So we conclude that $$ \frac{5x+4}{2x+3} = \frac 5 2 - \frac{-7/2}{2x+3}. $$ Then we have a constant, $-7/2$, in the numerator. $$ {} $$ – Michael Hardy Aug 19 '16 at 19:51