What is wrong with this factoring by completing the square?

Question

This is the problem and my attempt at solution:

$3x^2 + 2x - 1 = $

$3(x^2 + \frac{2}{3}x - \frac{1}{3}) = $

$3(x^2 + 2x + 1 - \frac{4}{3}x - \frac{4}{3}) = $

$3[(x + 1)^2 - \frac{4}{3}x - \frac{4}{3}] = $

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

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

$(x + 1)(3(x + 1) - 4) = $

$(x + 1)(3x - 1)$

However the textbook says this is the solution:

$(x + 1)(x - \frac{1}{3})$

Did I mess something up with the $3$ coefficient? I can't seem to notice what my mistake is.

Answer 1

expanding your result you will get $$3x^2+3x-x-1=3x^2+2x-1$$ it is the same term as in your exercise.

Answer 2

$(x+1)(3x+1)$ is the same as $3(x+1)(x+\frac13)$, because $3x+1=3(x+\frac13)$. Are you sure that the book doesn't say $3(x+1)(x+\frac13)$?