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)
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
Hello, brianbrian!
$\displaystyle \text{Express }\,\frac{3x^2+5x+4}{x^3+x^2+x+1}\,\text{ as a sum of partial fractions.}$
We have: .$\displaystyle \frac{3x^2+5x+4}{(x+1)(x^2+1)} \;=\;\frac{A}{x+1} + \frac{Bx}{x^2+1} + \frac{C}{x^2+1}$
. . . . . . . . . .$\displaystyle 3x^2+5x+4 \;=\;A(x^2+1) + Bx(x+1) + C(x+1)$
Let $\displaystyle x = \text{-}1: \;\;2 \;=\;2(A) + B(0) + C(0) \quad\Rightarrow\quad A \,=\,1$
Let $\displaystyle x = 0:\;\;4 \:=\:A + B(0) + C \quad\Rightarrow\quad 4 \:=\:1 + C \quad\Rightarrow\quad C \,=\,3$
Let $\displaystyle x = 1\!:\;\;12 \:=\:A(2) + B(2) + C(2) \quad\Rightarrow\quad 12 \:=\:2 + 2B + 6 \quad\Rightarrow\quad B \,=\,2$
Therefore: .$\displaystyle \frac{3x^2 + 5x + 4}{x^3+x^2+x+1} \;=\;\frac{1}{x+1} + \frac{2x}{x^2+1} + \frac{3}{x^2+1}$
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).
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: $\displaystyle \frac{Bx+ C}{x^2+ 1}= \frac{Bx}{x^2+ 1}+ \frac{C}{x^2+1}$.