TITLE: In order to factor we must find its zeros?
QUESTION [6 upvotes]: 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. :(

REPLY [2 votes]: 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 $$(x-3)(x-1+\sqrt2)(x-1-\sqrt2).$$