Let $f(x) = x^2 - 9x- 10$

We can state that $f(x) = (x + 1)(x-10)$ since I simply factored it.

The roots of this function is $-1$ and $10$. However, what is the relationship between a factored polynomial and its roots? Why can we assume this?

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    $(x-a)(x-b)=x^2-(a+b)x+ab$ – Vincent Law Aug 9 at 21:15
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    If there are non-trivial zero divisors (for example in the context of a 'field" or an 'integral domain' ('domain') - rational numbers or real numbers for the first, integers for the second) Alfred's answer is fine. If you were working with integers modulo $4$ you would have things like $2\times 2 \equiv 0$ or modulo $6$ you have $2\times 3 \equiv 0$ and you cannot assume that one of the individual factors must be zero. Ignore this comment if it is unhelpful, but if it intrigues you, there is much to explore. – Mark Bennet Aug 9 at 21:21
up vote 46 down vote accepted

It's not an assumption. It is a general fact that if $ab = 0$, then either $a = 0$ or $b = 0$. The roots of a polynomial are the numbers for which that polynomial evaluates to $0$, so to find the roots, it is enough to find the roots of the factors. So we factor everything as much as we can. In your case, this is a quadratic polynomial, so any factors are linear, and then it is obvious what the roots are.

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    @DRF No need to over complicate an answer to a “precalc” tagged question, imo. – YoTengoUnLCD Aug 10 at 7:41
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    @MartinBonner The reals are an integral domain. They are a very very special type of an integral domain given they are a field, and thus satisfy many more things than an integral domain must. An integral domain is just a commutative ring with no zero divisors. Fields get no zero divisors but they also get inverses which is much more. – DRF Aug 10 at 14:59
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    @MartinBonner "Integral" in "integral domain" has nothing to do with integration. Its meaning is closer to (but not equivalent to) "some sort of integer." Integral domains are a type of abstract-algebraic object that, in some sense, have all the "important" properties of the integers (and maybe more properties, but not less). Specific definitions can be found elsewhere, but colloquially this explains the name. The real numbers are an integral domain because they have all of the "important" properties of the integers and also a lot of other unrelated properties. – probably_someone Aug 10 at 15:38
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    @MartinBonner: An example of what is not an integral domain is the ring of $n \times n$ real matrices. There are nonzero $n \times n$ matrices $A,B$ such that $AB$ is the zero matrix. Another example is the ring of integers modulo $6$, because $2·3 \equiv 0 \pmod{6}$. – user21820 Aug 10 at 17:04
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    @probably_someone Just for some etymological backup: I found a paper from 1908 that defines an integral domain as a subring of a "domain" (a field). Integers used to be known as integral numbers, and the integral numbers in the domain of rationals form an integral domain, in this old language. Some authors even use "domain" to mean a subring of a field (which is equivalent to saying there are no zerodivisors). – Kyle Miller Aug 10 at 20:50

Alfred Yerger explains why factoring gives roots, but that does not explain the converse: why do roots give factors?

Let $p(x)$ be a polynomial with a root $c$. By long-division of $p(x)$ by $x-c$, we can write $p(x)=q(x)(x-c)+r$, where $r$ is the remainder (a constant since $x-c$ is a linear polynomial). Now, by substituting $x=c$, we get $p(c)=q(c)(c-c)+r$. Since $p(c)=0$ and $c-c=0$, we have $0=r$. That is, there was no remainder after all! Hence, the result of the long division is that $p(x)=q(x)(x-c)$. In other words, if we factor out $x-c$ from $p(x)$, we are left with a polynomial $q(x)$.

Something that is easy to miss with this is that in fact, $p(c)=q(c)(c-c)+r$. That is, $p(c)=r$, whether or not $c$ is actually a root. Thus, the long division will always give $p(x)=q(x)(x-c)+p(c)$. This is the basis of a computational technique called synthetic division which gives both $q(x)$ and $p(c)$. And, it takes about as much effort to calculate both as it does to calculate either on its own.

A polynomial of degree $d$ has exactly $d$ roots (Fundamental Theorem of Algebra), which can be complex and/or multiple.

If you factor it in $n$ factors, the factors are also polynomial and their degrees are such that $d=d_1+d_2+\cdots d_n$. This guarantees that no root "gets lost". On the other hand, any root of a factor is perforce a root of the initial polynomial.

Hence the roots of a polynomial are exactly the roots of the factors. The reason we factor is that the roots of the factors can be easier to find than those of the original polynomial.

In the case of polynomials with real coefficients, it can be more convenient to factor in first degree binomials for the real roots and second degree trinomials for the conjugate pairs. E.g. $x^3+2x^2+2x+1=(x+1)(x^2+x+1)$.

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