# rational function as a sum of partial fractions

• Jan 30th 2013, 10:13 AM
brianbrian
rational function as a sum of partial fractions

Express the rational function as a sum of partial fractions: f(x) = (3x^2+5x+4)\(x^3+x^2+x+1)
• Jan 30th 2013, 10:39 AM
Plato
Re: rational function as a sum of partial fractions
Quote:

Originally Posted by brianbrian
Express the rational function as a sum of partial fractions: f(x) = (3x^2+5x+4)\(x^3+x^2+x+1)

Have a look at this.
• Jan 30th 2013, 10:46 AM
brianbrian
Re: rational function as a sum of partial fractions
huh, thanks :) but i guess that teacher want me to do this with this A,B,C thing, so how can I compromise that result with this A,B,C (i hope you know what i am talking about, its hard to explain for me)
as ABC thing I mean situation when partial consists of letter in nominator and some (numbers with x) in denominator ;)
• Jan 30th 2013, 12:38 PM
Soroban
Re: rational function as a sum of partial fractions
Hello, brianbrian!

Quote:

$\text{Express }\,\frac{3x^2+5x+4}{x^3+x^2+x+1}\,\text{ as a sum of partial fractions.}$

We have: . $\frac{3x^2+5x+4}{(x+1)(x^2+1)} \;=\;\frac{A}{x+1} + \frac{Bx}{x^2+1} + \frac{C}{x^2+1}$

. . . . . . . . . . $3x^2+5x+4 \;=\;A(x^2+1) + Bx(x+1) + C(x+1)$

Let $x = \text{-}1: \;\;2 \;=\;2(A) + B(0) + C(0) \quad\Rightarrow\quad A \,=\,1$

Let $x = 0:\;\;4 \:=\:A + B(0) + C \quad\Rightarrow\quad 4 \:=\:1 + C \quad\Rightarrow\quad C \,=\,3$

Let $x = 1\!:\;\;12 \:=\:A(2) + B(2) + C(2) \quad\Rightarrow\quad 12 \:=\:2 + 2B + 6 \quad\Rightarrow\quad B \,=\,2$

Therefore: . $\frac{3x^2 + 5x + 4}{x^3+x^2+x+1} \;=\;\frac{1}{x+1} + \frac{2x}{x^2+1} + \frac{3}{x^2+1}$
• Jan 30th 2013, 12:58 PM
brianbrian
Re: rational function as a sum of partial fractions
Great! Thanks so much. Unfortunatly small question appeared. Why there is a x next to B? I think that i understand the whole idea now, but i can't get why this x stands next to B (and why B, not A or C).
• Jan 30th 2013, 02:46 PM
HallsofIvy
Re: rational function as a sum of partial fractions
Any polynomial can be factored (at least theoretically- it might be extremely difficult to find the coefficients) into linear or quadratic factors (with real coefficients). Since we want "common fractions", we need the degree of the numerator to be less than that of the denominator. So the numerator of a fraction having a linear denominator is a number and the numerator of a fraction having a quadratic denominator is linear. Soroban separated the fraction having linear numerator and quadratic denominator into two fractions: $\frac{Bx+ C}{x^2+ 1}= \frac{Bx}{x^2+ 1}+ \frac{C}{x^2+1}$.