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. :(

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    $\begingroup$ If $(x - a)$ is a factor, then $f(a) = 0$, and vice versa $\endgroup$ – The Chaz 2.0 May 6 '13 at 18:57
  • 1
    $\begingroup$ You can proceed to factor $(x-1)^2-2$ using $a^2-b^2=(a+b)(a-b)$ with $a=x-1$. (What is $b$?) As was pointed out by @TheChaz2.0, factoring and finding zeroes is equivalent. $\endgroup$ – gt6989b May 6 '13 at 18:59
  • $\begingroup$ I just edited my question to provide more information on exactly what my confusion is about. $\endgroup$ – Anthony May 6 '13 at 19:41
  • $\begingroup$ To find the roots of the quadratic you could just use the quadratic formula. Given $ax^2+bx+c$, your solutions will be $\displaystyle\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. $\endgroup$ – Brian Silva May 6 '13 at 20:53
  • $\begingroup$ It is usually wrong to say that to do $P$ you must first do $Q.$ There is usually more than one way to do something in math. What is OK to say is, I have a method for doing $P,$ and the first step is to do $Q.$ $\endgroup$ – David K Sep 27 '18 at 13:31

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 $$(x-3)(x-1+\sqrt2)(x-1-\sqrt2).$$

  • $\begingroup$ I'm at a cafe which is closing soon, but this seems to be the answer I'm looking for; that is, it seems to explain why finding zero is important for factoring but I'll need to study it so that it sinks in. I'll read this in more detail either tonight or tomorrow morning and let you know if it makes a dent in my potato brain. $\endgroup$ – Anthony May 6 '13 at 21:32

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 $P(x)=(x-3)(x-(1+\sqrt2))(x-(1-\sqrt2))$


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 $$P(x) = (x-3)(x-x_1)(x-x_2)$$


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.

  • $\begingroup$ It was $x^2-2x-1$, not $x^2-2x+1$. $\endgroup$ – MJD May 6 '13 at 19:41

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