rational function as a sum of partial fractions

**brianbrian** · January 30th 2013, 10:13 AM

Please solve this

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

**Plato** · January 30th 2013, 10:39 AM

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.

**brianbrian** · January 30th 2013, 10:46 AM

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

**Soroban** · January 30th 2013, 12:38 PM

Hello, brianbrian!

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

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: . $\frac{3x^2 + 5x + 4}{x^3+x^2+x+1} \;=\;\frac{1}{x+1} + \frac{2x}{x^2+1} + \frac{3}{x^2+1}$

**brianbrian** · January 30th 2013, 12:58 PM

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).

**HallsofIvy** · January 30th 2013, 02:46 PM

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: $\frac{Bx+ C}{x^2+ 1}= \frac{Bx}{x^2+ 1}+ \frac{C}{x^2+1}$ .

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