How do you long divide $(x-3)/(x^3+2x^2+x)$ ?

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Answer 1 · 2017-03-29T11:18:58

$-3/x +3/(x + 1) + 4/(x + 1)^2$

I think you want partial fractions.
Factorize below: $x(x^2 + 2x + 1) = x(x + 1)^2$

$frac{x-3}{x^3 + 2x^2 + x} = A/x +B/(x + 1) + C/(x + 1)^2$
$exists A,B,C in RR$

$x - 3 = A(x + 1)^2 +Bx(x+1) + Cx$

$x^2 : 0 = A + B$
$x^1: 1 = 2A + B + C$
$x^0: -3 = A Rightarrow B = +3$

$1 = 2(-3) + 3 + C Rightarrow C = 4$

How do you long divide (x-3)/(x^3+2x^2+x)(x-3)/(x^3+2x^2+x)?