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The remainder on dividing by a linear polynomial is the value there, so a root is a factor seen from the other side — and a polynomial of degree $n$ over a field has at most $n$ of them.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to use the remainder theorem to find a remainder without dividing, decide whether a linear polynomial is a factor by one substitution, search a small field exhaustively for roots, prove the factor theorem from a single division, state the bound on the number of roots and identify the hypothesis it needs, and explain why the bound fails over a ring with zero divisors.
Dividing by a linear polynomial is the simplest case of the algorithm from the last lesson, and it is the one worth doing in your head. The remainder has to be a constant, and substituting the right number identifies which constant it is — from which the whole relationship between roots and factors follows.
A root of $f$ in a field $F$ is an $a \in F$ with $f(a) = 0$. The remainder theorem says the remainder on dividing $f$ by $x - a$ is $f(a)$. The factor theorem says $x - a$ divides $f$ exactly when $f(a) = 0$. The multiplicity of a root $a$ is the largest $k$ with $(x - a)^{k}$ dividing $f$.
Divide $f$ by $x - a$. The remainder has degree below $1$, so it is a constant $r$:
$$f(x) = q(x)(x - a) + r.$$
Substituting $x = a$ kills the first term whatever $q$ is, so $r = f(a)$. That is the remainder theorem, and the factor theorem is the same equation read twice:
$$x - a \mid f \iff f(a) = 0.$$
Counting roots. Suppose $f$ has degree $n \ge 1$ over a field and $a_1$ is a root. Then $f = (x - a_1)g$ with $\deg g = n - 1$. If $a_2$ is another root, then $0 = f(a_2) = (a_2 - a_1)g(a_2)$, and since $a_2 - a_1 \ne 0$ in a field, $g(a_2) = 0$: the new root is a root of $g$. Repeating, each root drops the degree by one, so
> a non-zero polynomial of degree $n$ over a field has at most $n$ roots.
Counted with multiplicity the same bound holds, and the two statements come apart exactly at repeated roots: $(x - 2)^{2}$ has one root and two linear factors.
The hypothesis is doing work. The step that needs it is a product is zero only when a factor is — the integral domain condition. Over $\mathbb{Z}_8$, which has zero divisors, $x^{2} - 1$ has the four roots $1, 3, 5, 7$. The factor theorem itself survives there (the divisor $x - a$ is monic, so no inverse is needed), but the counting does not.
At most, not exactly. $x^{2} - 2$ has no rational root and $x^{2} + 1$ has no real one. That every polynomial of degree $n$ over $\mathbb{C}$ has exactly $n$ roots with multiplicity is the fundamental theorem of algebra — a fact about $\mathbb{C}$, not about polynomials.
Another way: picture
Think of a root as a hook that a linear factor hangs on. Each hook takes one factor and one unit of degree, so a polynomial of degree $n$ has at most $n$ hooks. What the picture makes clear is where zero divisors break it: they let a product hang on a hook without either factor doing so, and then the count means nothing.
Another way: steps
To find the roots of $f$ over a small field: 1. Substitute every element of the field and reduce; the zeros are the roots. 2. For each root $a$, divide by $x - a$ to get a polynomial of lower degree. 3. Repeat on the quotient, looking for repeated roots as well. 4. Stop when the degree is $0$, or when the remaining factor has no root. 5. Check: the number of roots never exceeds the original degree.
Testing a factor cheaply. To decide whether $x - a$ divides $f$, evaluate rather than divide: one substitution against a whole long division. For a polynomial with integer coefficients and leading coefficient $1$, any integer root divides the constant term, so the candidates are the divisors of that term — a short list.
Searching a finite field exhaustively. Over $\mathbb{Z}_p$ there are only $p$ elements, so substituting all of them settles the question of roots completely. That is what makes irreducibility decidable over a finite field, and it is the method of the next lesson.
A warning. Over a finite field a polynomial is not the same thing as the function it defines. Over $\mathbb{Z}_5$, the polynomial $x^{5} - x$ is not the zero polynomial — it has degree $5$ and non-zero coefficients — yet it vanishes at every one of the five elements, because $a^{5} = a$ there by Fermat's little theorem. So two different polynomials can define the same function.
This matters for the bound. $x^{5} - x$ has five roots and degree five, which is consistent; what would be inconsistent is a non-zero polynomial of degree less than five vanishing everywhere, and no such thing exists. The bound is about polynomials as formal expressions, and everything in this unit treats them that way. Over an infinite field the distinction disappears — two polynomials agreeing as functions are equal, because their difference would have infinitely many roots.
*Reading at most $n$ as exactly $n$.* $x^{2} + 1$ has no real root at all. The bound limits; it does not produce.
Counting a repeated root twice when listing roots. $(x - 2)^{2}$ has one root, of multiplicity two. Both counts are useful; say which one you mean.
Using the bound without a domain. $x^{2} - 1$ has four roots modulo $8$. The counting argument needs a product to be zero only when a factor is.
Thinking the factor theorem fails over a ring with zero divisors. It does not: $x - a$ is monic, so it can be divided by anywhere. It is the counting that fails.
Identifying a polynomial with its function over a finite field. $x^{p} - x$ vanishes everywhere on $\mathbb{Z}_p$ and is not the zero polynomial.
Is $x - 3$ a factor of $f(x) = x^{3} - 2x^{2} - 5x + 6$ over the rationals?
The question.
Evaluate: $f(3) = 27 - 18 - 15 + 6 = 0$.
One substitution.
So yes, and dividing gives $f = (x - 3)(x^{2} + x - 2) = (x - 3)(x + 2)(x - 1)$: three roots for a cubic, which is the bound attained.
Factor, then continue.
Take $x^{2} + 2$ over $\mathbb{Z}_5$ and substitute all five elements: $2, 3, 1, 1, 3$.
A complete search.
None is zero, so there is no root, and by the factor theorem no linear factor.
No roots, no linear factors.
A quadratic with no linear factor cannot factor at all, so $x^{2} + 2$ is irreducible over $\mathbb{Z}_5$ — and the quotient by it is a field with $25$ elements.
Which is the last lesson of the course.
At most four, since the degree is four and each root peels off a linear factor.
The bound.
Over the rationals it has two, $1$ and $-1$, since $x^{4} - 1 = (x^{2} - 1)(x^{2} + 1)$ and the second factor has no rational root.
The bound is not attained.
Over the complex numbers it has four: $1, -1, i, -i$. Same polynomial, same bound, different fields — and the count depends on the field, not on the polynomial alone.
Work over the integers modulo $5$, which is a field. Evaluate $x^2 + 2$ at each element.
| Its value | |
|---|---|
| at 0 | |
| at 1 | |
| at 2 | |
| at 3 | |
| at 4 |
Let $f(x) = x^{2} + 1$ over the rationals. What is the remainder on dividing $f$ by $x - 3$?
Answer:
Build the proof that $x - a$ divides $f$ exactly when $f(a) = 0$.
This task has no paper form; do it on a device.
Match each polynomial over the rationals to the number of rational roots it has.
| no rational roots | one rational root | two rational roots | three rational roots | four rational roots | |
|---|---|---|---|---|---|
| $x^{2} - 4$ | |||||
| $x^{2} - 2$ | |||||
| $x^{2} - 4x + 4$ | |||||
| $x^{3} - x$ |
Work over the integers modulo $5$. Is $x - 2$ a factor of $x^2 + 4$?
The polynomial $x^{2} - 1$ has four roots modulo $8$, namely $1, 3, 5$ and $7$. Select every statement that correctly accounts for this.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Work over the integers modulo $5$, which is a field. Evaluate $x^2 + 2$ at each element.
| Its value | |
|---|---|
| at 0 | |
| at 1 | |
| at 2 | |
| at 3 | |
| at 4 |
You can find the remainder on dividing by $x - a$ without dividing, and say how many roots a polynomial of given degree can have. Say in your own words why the bound needs the coefficients to have no zero divisors. Next: irreducibility, where having no root is only sometimes enough.
9. Your turn: how many roots can $x^{4} - 1$ have over a field?, step 3