7. Polynomial identities, roots and rational-root arguments

Learning goals and algebraic setting

Distinguish a polynomial identity from equality at a few values, use factor multiplicities, and prove restrictions on rational roots. Work over a field $F$, here $\mathbb Q$, $\mathbb R$ or $\mathbb C$. A polynomial is a finite formal expression $a_0+a_1x+\cdots+a_nx^n$ with coefficients in $F$. Its degree is the highest index with nonzero coefficient; leave the zero polynomial's degree undefined.

The division algorithm over a field gives $p=dq+r$ for $d\ne0$ and $r=0$ or $\deg r<\deg d$, by successively cancelling leading coefficients. Division by $x-a$ yields the remainder $p(a)$. Hence $a$ is a root exactly when $x-a$ divides $p$. A root has multiplicity $m$ when $(x-a)^m$ divides $p$ but $(x-a)^{m+1}$ does not.

Theorem: root bounds and interpolation uniqueness

A nonzero polynomial of degree $n$ over a field has at most $n$ distinct roots. Consequently, polynomials of degree at most $n$ agreeing at $n+1$ distinct points are identical.

Proof. By induction on $n$, a nonzero constant has no roots. If a degree-$n$ polynomial has a root $a$, factor $p=(x-a)q$, with $q$ of degree $n-1$. Every other root $b\ne a$ satisfies $0=(b-a)q(b)$, so $q(b)=0$ because $b-a\ne0$ is invertible. The induction hypothesis gives at most $n-1$ others. For the second assertion, subtract the two polynomials. A nonzero difference of degree at most $n$ could not have $n+1$ distinct roots, so the difference is zero.

Theorem: rational-root restriction

Let $p(x)=a_nx^n+\cdots+a_0$ have integer coefficients, $a_n\ne0$, and let $u/v$ be a rational root in lowest terms with $v>0$. Then $u\mid a_0$ and $v\mid a_n$.

Proof. Multiply $p(u/v)=0$ by $v^n$. The resulting equality shows that $u$ divides $a_0v^n$ and $v$ divides $a_nu^n$. Since $\gcd(u,v)=1$, Euclid's lemma (or the coprime cancellation consequence of Bezout) gives $u\mid a_0$ and $v\mid a_n$. If $u=0$, roothood forces $a_0=0$; the divisibility statement is interpreted as $0\mid0$, true under the definition. Often one first removes powers of $x$ and applies the rule only to nonzero candidate roots.

Polynomial division: termination, uniqueness and evaluation

Let $F$ be one of the fields $\mathbb{Q}$, $\mathbb{R}$ or $\mathbb{C}$, and use finite formal polynomials in $F[t]$. Degree is used only for nonzero polynomials; $\operatorname{lc}(u)$ denotes the leading coefficient of a nonzero polynomial $u$. If $u$ and $v$ are nonzero, their leading coefficients have a nonzero product, since every nonzero field element is invertible. Thus $\deg(uv)=\deg u+\deg v$. A nonzero sum or difference has degree at most the larger input degree; a zero result is handled separately.

Division theorem and complete proof. For $p,d$ in $F[t]$ with $d$ nonzero, there is a unique pair $q,r$ with $p=dq+r$ and either $r=0$ or $\deg r<\deg d$. Put $m=\deg d$, initially $q=0$ and $r=p$. If $r$ is nonzero with degree at least $m$, put $k=\deg r-m$ and $c=\operatorname{lc}(r)/\operatorname{lc}(d)$. Replace $q$ by $q+ct^k$ and $r$ by $r-ct^k d$. The identity $p=dq+r$ is preserved. The leading terms of the two expressions in the subtraction agree, so the new $r$ is zero or has strictly smaller degree. All nonzero residual degrees are nonnegative integers; there can be only finitely many such steps. At termination the required remainder condition holds. This also covers constant $d$, when the final remainder must be zero, and $p=0$, which requires no step.

For uniqueness, suppose $dq+r=dq'+r'$, with both remainder conditions. If $q-q'$ were nonzero, the nonzero polynomial $d(q-q')$ would have degree at least $m$. But it equals $r'-r$, which is zero or has degree less than $m$, a contradiction. Hence $q=q'$, and the identity then gives $r=r'$.

For $a\in F$, evaluation at $a$ preserves sums and products: distribute the finite sums of coefficients, and use $a^{i+j}=a^i a^j$ in each product term. Divide by $t-a$. Its remainder is a constant $r$, possibly zero, so evaluation gives $p(a)=r$. Consequently $p(a)=0$ if and only if $t-a$ divides $p$. These conclusions do not assume that a polynomial has a root, or that the field is algebraically closed.

Worked examples

Irreducible over one field but not another. The polynomial $x^2-2$ has no rational root by the rational-root rule (only $\pm1,\pm2$ need testing). A degree-two polynomial over a field is reducible exactly when it has a root there: any nontrivial factorization must use two linear factors. Thus $x^2-2$ is irreducible over $\mathbb Q$, but over $\mathbb R$ it factors as $(x-\sqrt2)(x+\sqrt2)$. The coefficient field is part of the claim.

Reconstructing a quadratic. Seek $p(x)=ax^2+bx+c$ with $p(0)=1,p(1)=0,p(2)=3$. The first condition gives $c=1$; the other two give $a+b=-1$ and $4a+2b=2$. Thus $a=2,b=-3$, so $p=2x^2-3x+1$. Substitution proves existence; the root-bound theorem proves uniqueness among degree-at-most-two polynomials.

Odd real degree: the leading-term sign and the stated IVT input

Let $p(x)=a_nx^n+\sum_{j=0}^{n-1}a_jx^j$ be a real polynomial of positive odd degree $n$, with $a_n$ nonzero. Write $A=\sum_{j=0}^{n-1}|a_j|$. For $|x|\ge1$ the lower terms have total absolute value at most $A|x|^{n-1}$. Choose $R>\max(1,2A/|a_n|)$. At $x=R$ and $x=-R$ this error is strictly less than $|a_nx^n|/2$, so $p(x)$ has the sign of $a_nx^n$ at both endpoints. Odd $n$ makes those signs opposite. A polynomial is continuous by the finite sum/product rules, and the intermediate value theorem on $[-R,R]$ supplies a zero. The IVT is an explicit input of this exercise, not a theorem about complex ordering or a claim of proving the fundamental theorem of algebra.

Graded practice

1. Foundation. Divide $x^3-1$ by $x-1$ and identify its real roots.
2. Core. List all possible nonzero rational roots of $2x^3-3x+1$, then factor it over $\mathbb Q$.
3. Core. Prove that an odd-degree real polynomial has at least one real root, using continuity and the intermediate value theorem.
4. Proof. Can a nonzero real polynomial satisfy $p(n)=0$ for every positive integer $n$? Can a nonzero function do so? Explain the distinction.

Full solutions

1. $x^3-1=(x-1)(x^2+x+1)$. The quadratic discriminant is $1-4=-3$, so its real root set is empty. The only real root is one.
2. The candidate list is $\pm1,\pm1/2$. Substitution gives values $0,2,-1/4,9/4$, respectively, so only $1$ is a rational root. Division by $x-1$ yields $2x^2+2x-1$, and thus $2x^3-3x+1=(x-1)(2x^2+2x-1)$, whose quadratic has roots $(-1\pm\sqrt3)/2$, irrational. No further factorization exists over $\mathbb Q$.
3. The leading term dominates at large positive and negative inputs. For odd degree its signs at the two ends are opposite, whether its leading coefficient is positive or negative. Choose finite endpoints with those opposite signs. Continuity and the intermediate value theorem give a zero between them.
4. No nonzero polynomial can have infinitely many distinct roots, by the degree bound. A nonzero function can: $f(x)=\sin(\pi x)$ vanishes at every integer but, for example, $f(1/2)=1$. Polynomial rigidity does not extend to arbitrary continuous or smooth functions.

Summary and term definitions

A polynomial is irreducible over a field when it admits no factorization into two positive-degree polynomials over that field. A polynomial identity holds as an equality of coefficient expressions, not just at selected inputs. Interpolation constructs a polynomial matching specified data. Root tests supply candidates, not guaranteed roots; substitute every candidate and state the coefficient field before claiming a factorization complete.