Something different about partial factorization with wolfram alpha

Question

I was factorizing $$f(x) = \frac{4x^2+7x+2}{4x^3+4x^2+2x}$$
I approached this like below
$$f(x) = \frac{a}{2x} + \frac{bx+c}{2x^2+2x+1}$$ and got a = 2, b = 0, c = $\frac{3}{2}$. So, f(x) becomes

$$f(x) = \frac{1}{x} + \frac{3}{2x^2+2x+1}$$ (multiplied 2 to c because I reduced $\frac{2}{2x}$)

However, when I run this in wolfram alpha I get something like below,
Wolfram screen shot
I can't find an explanation for the denominators splitting into $2x$ and $ 2(2x^2+2x+1)$ where did the 2 in the 2nd denominator come from?

Answer 1

A numerical factor in the left-hand denominator can appear on the right-hand side in a number of ways.

\begin{align*} \frac{1}{2} \cdot \frac{4x^2+7x+2}{x(2x^2+2x+1)} &= \frac{1}{2} \left( \frac{a}{x} + \frac{bx+c}{2x^2+2x+1} \right) \\ &= \frac{a}{2x} + \frac{bx+c}{2(2x^2+2x+1)} \text{,} \end{align*} \begin{align*} \frac{1}{2} \cdot \frac{4x^2+7x+2}{x(2x^2+2x+1)} &= \frac{1}{2} \left( \frac{a}{x} + \frac{bx+c}{2x^2+2x+1} \right) \\ &= \frac{a/2}{x} + \frac{(b/2)x+(c/2)}{2x^2+2x+1} \text{,} \end{align*} or even a combination such as in your version \begin{align*} \frac{1}{2} \cdot \frac{4x^2+7x+2}{x(2x^2+2x+1)} &= \frac{1}{2} \left( \frac{a}{x} + \frac{bx+c}{2x^2+2x+1} \right) \\ &= \frac{a}{2x} + \frac{(b/2)x+(c/2)}{2x^2+2x+1} \text{.} \end{align*}

Answer 2

The $\,2\,$ in the denominator is a numerical common factor and should be removed before separating into partial fractions. In other words, the steps should be

$$ \begin{align} \frac{4x^2+7x+2}{4x^3+4x^2+2x} &= \frac{1}{2} \left[\frac{4x^2+7x+2}{2x^3+2x^2+x}\right]\\[1ex] &= \frac{1}{2} \left[\frac{a}{x} + \frac{bx+c}{2x^2+2x+1}\right]\\[1ex] &= \frac{1}{2} \left[\frac{2}{x} + \frac{3}{2x^2+2x+1}\right]\\[1ex] &= \frac{1}{x} + \frac{3}{2(2x^2+2x+1)}\\[1ex] \end{align} $$