Finding all the rational roots of $25x^3+25x^2-x-1$ Finding all the rational roots of $25x^3+25x^2-x-1$.

So, I saw right away that $-1$ was a root. I then used synthetic division to factor this as:

$(25x^2-1)(x+1)$. 

Then I found the roots to $25x^2-1$ as $\frac{1}{5},\frac{-1}{5}$. 

I've been telling my precalculus class that once you know the roots you know the factors, and if you know the factors you know the roots. However, in this case the original polynomial does not equal $(x+1)(x-\frac{1}{5})(x+\frac{1}{5})$. Can somebody tell me exactly what went wrong here? Thanks!!
The thing is, the fundamental theorem of algebra admits a leading coefficient. Let's be more explicit. Say a polynomial $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots a_1 x + a_0$ has the roots $r_1, \cdots, r_n$, which may be complex numbers and not necessarily distinct. The leading coefficient, $a_n$, is assumed nonzero.

Then we can write $f$ as a product of linear factors in terms of its roots and its leading coefficient:

$$f(x) = a_n (x-r_1)(x-r_2)\cdots (x-r_n)$$

This is what you forgot, the leading coefficient. The roots are not sufficient to uniquely determine the polynomial on their own - you can easily see that, however you vary it, the roots would remain the same since $a_n$ is just a constant factor.



With this in mind, augmenting your procedure in this example is quite simple. Find the roots of $f(x)$ as you have, and note that the leading coefficient is $25$. Then

$$f(x) = 25(x + 1)\left( x - \frac 1 5 \right)\left( x + \frac 1 5 \right)$$

Of course, that $25$ can be distributed into the rightmost two factors for a nicer presentation if you so choose. 

$$f(x) = (x + 1) \cdot 5 \cdot \left( x - \frac 1 5 \right) \cdot 5 \cdot \left( x + \frac 1 5 \right) = (x+1)(5x-1)(5x+1)$$

However this is purely cosmetic and not at all necessary.
Simple factorization....
$$
25x^3+25x^2-x-1=25x^2(x+1)-1(x+1)=(x+1)(25x^2-1))
$$
$$
=(x+1)(5x-1)(5x+1)
$$