In order to factor we must find its zeros? I am self-learning precalc (Precalculus Demystified) and found the following problem on page 170 :
Completely factor the the polynomial. $P(x) = x^3 - 5x^2 + 5x + 3; c = 3$ is a zero.
Since $c = 3$ is a zero, I know $x - 3$ is a factor (and remainder is zero). So I go ahead and do the division and end up with the following :
$(x - 3)(x^2 - 2x - 1)$
However, this needs to be factored even further, so I do the square and end up with :
$(x - 3)((x - 1)^2 - 2)$
Now I'm scratching my head because it doesn't look like the nice factors I'm used to seeing, so I check my progress in the book and they say the following :
"In order to factor $x^2 - 2x - 1$, we must first find its zeros"
Now I understand that in order to find the intercepts on a graph I must find its zeros, but what does finding its zeros have to do with factoring this equation? I'm completely lost at this point because though I can take it at face value and proceed, I need to understand the why before I feel comfortable plowing ahead or I'm going to dig myself into an even deeper pool of confusion. :(
Any thoughts? Thanks in advance!
Update : Due to the answers I'm getting, I need to clarify my question to pinpoint my actual confusion.
Thank you for your quick replies! I'm so sorry but I'm a little slow in understanding the implication. My problem is that I don't understand how factoring and finding zeros is related. For me, factoring is finding what are the root divisors of a number or equation. For example, 3 is a factor of 9 because 3 * 3 = 9. When I look for factors of 9, I don't think about finding zeros... I don't even know what that means! I just look to see what multiplied by what gives me 9. In an equation such as x^2 + 2xy + y^2, I understand how to factor this to (x + y)^2... this I understand the why of and how to do it. But in all the times I have done it, never once have I thought about "finding the zero" of anything nor have I approached it thinking about zeros at all. This may betray an immense ignorance of some basic understanding on my part, but this is why I'm asking this question... I would love to know what I'm missing and what zeros have to do with anything. For finding the x/y intercepts, yes, I understand how zero relates... for factoring I don't. :(
 A: Let me first explain why finding the roots helps us to factor. If $f(x)$ and $g(x)$ are polynomials with $g(x)$ not the zero polynomial, then polynomial long division tells us that there exist polynomials $q(x)$ and $r(x)$ such that $f(x)=g(x)q(x)+r(x),$ where $r$ is the zero polynomial, or has degree less than that of $g(x)$.
In particular, let's consider the case where $g(x)=x-c$ for some constant $c$. Then regardless of what the polynomial $f(x)$ is, the remainder polynomial $r(x)$ will be a constant polynomial, since it's either the zero polynomial or it's a non-zero polynomial of degree less than $1$ (so degree $0$). In other words, for any polynomial $f(x)$ and any constant $c$, there exist a polynomial $q(x)$ and a constant $r$ such that $$f(x)=(x-c)q(x)+r.$$ In particular, $$\begin{align}r &= 0\cdot q(c)+r\\ &= (c-c)q(c)+r\\ &= f(c)\end{align}$$ so for all polynomials $f(x)$ and all constants $c$, there is a polynomial $q(x)$ such that $$f(x)=(x-c)q(x)+f(c).$$ But this means that $x-c$ is a factor of $f(x)$ if and only if $f(c)=0$. (Do you see why that follows from the above equation?) This is why finding zeroes is effectively the same thing as finding linear factors, which helps us to completely factor a polynomial.

Now, to your specific example. You've already rewritten your polynomial as $$(x-3)\bigl((x-1)^2-2\bigr).$$ Now, note that we can rewrite $$(x-1)^2-2=(x-1)^2-(\sqrt2)^2,$$ so by the difference of squares formula $a^2-b^2=(a+b)(a-b)$, we can write $$(x-1)^2-2 = (x-1+\sqrt2)(x-1-\sqrt2),$$ and so your polynomial in completely factored form is $$(x-3)(x-1+\sqrt2)(x-1-\sqrt2).$$
A: You've got it right up to $(x−3)((x−1)^2−2)$
From this point, you must search for the solution of $(x-1)^2=2$.
Then you get the zeros : $x_1=3, x_2=1+\sqrt2 $ and $x_3=1-\sqrt2$
And you can write your function as $P(x)=(x-3)(x-(1+\sqrt2))(x-(1-\sqrt2))$
A: Maybe an easier example will help. 

Completely factor $f(x) = x^3 +2x^2 -11x -12$, using the fact that f(3) = 0.

So you can factor out $(x - 3)$ or use synthetic division to get the remainder. In any case, you'll have $f(x) = (x - 3)(x^2 + 5x + 4)$ - which can be further factored into: $$(x-3)(x+1)(x+4)$$
Your example is a little trickier because $x^2 -2x -1$ cannot be factored over the rationals. So you basically find the roots $x_1,x_2$ using the quadratic formula, then write $$P(x) = (x-3)(x-x_1)(x-x_2)$$
A: To find the '$o$'s' you equate the function to $0$, so $f(x)=0$. However, if we can first write $f(x)$ in several factors, we can make use of the fact that for a product to be equal to $0$, one (or more) of its factors has to be $0$.
For example let's take $x^2 - 2x + 1$ that you wrote in your problem statement. We can factor this as $(x-1)(x-1)$. Now we want to find the '$0$'s' so $x^{2} - 2x + 1=0$ or equivalently $(x-1)(x-1)=0$. Now we make use of the rule that I just mentioned, so we know that the first $(x-1)$ factor should be $0$ and/or the other $(x-1)$ factor should be equal to $0$. Hence, $x-1=0 -> x=1$. We can now observe that the numbers in the factors belong to the '$0$'s'of the function.
