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For a field $F$, polynomial division makes the one-variable polynomial ring $F[X]$ a Euclidean domain with Euclidean function equal to the polynomial degree. For a divisor $g\ne 0$, subtract a suitable scalar multiple of $X^{\mathrm{deg}f-\mathrm{deg}g}g$ to cancel the leading coefficient of the current remainder. Its degree strictly decreases. Iteration terminates at $f=qg+r$, where $r=0$ or $\mathrm{deg}r<\mathrm{deg}g$. Division by the leading coefficient is valid because it is nonzero in a field.

In polynomial division $q=Qs+r$ over a field, the polynomial quotient is the polynomial $s$. The polynomial remainder $r$ has smaller polynomial degree than the nonzero divisor $Q$ unless it is zero.

In the notation $f=gq+r$, the polynomial quotient is instead named $q$. It is a polynomial determined by the divisor and the polynomial degree bound. The pointwise ratio of the associated polynomial functions satisfiesFor the worked example, $f(0)/g(0)=6/(-2)=-3$, while $q(0)=1$; the correction is $r(0)/g(0)=8/(-2)=-4$. Thus a polynomial quotient need not equal the rational function $f/g$. At $x=1$ and $x=-2$, the pointwise ratio is undefined, but the polynomial quotient still has perfectly well-defined values. When $r=0$, the divisor is a polynomial factor of $f$ and the ratio agrees with $q$ wherever $g(x)\ne 0$.

The polynomial remainder is the polynomial $r$ in polynomial division $q=Qs+r$. Division by $X-a$ has a constant polynomial remainder $q(a)$, as expressed by the factor theorem.

For any identity $f=gq+r$, polynomial evaluation at a root of a polynomial $a$ of $g$ givesThis substitution requires no division by $g(a)$. In the worked example, polynomial factorization gives $g=(X-1)(X+2)$, so its two roots of a polynomial are $1$ and $-2$. The cubic polynomial and its linear polynomial remainder agree there:A cubic expression has been compressed into a straight line carrying exactly these two values.

Conversely, those two values determine the linear polynomial remainder by polynomial interpolation. Write $r=AX+B$; then $A+B=6$ and $-2A+B=12$, giving $A=-2$ and $B=8$. The polynomial degree bound is essential: adding any multiple of $g$ preserves the two values, but generally raises the polynomial degree. The plot also shows agreement at $x=-1/2$, where $q=0$ rather than $g=0$. Agreement of $f$ and $r$ at a point therefore does not imply that the point is a root of a polynomial of the divisor.

Take a different real matrix,Its only eigenvalue is $2$, with algebraic multiplicity two. However, the polynomial $q(t)=t-2$, which vanishes at that eigenvalue, gives $q(B)=N\ne 0$. Merely checking values at eigenvalues cannot prove a matrix identity. The full characteristic polynomial does work:The matrix is a Jordan block, and its minimal polynomial is also $(t-2{)}^2$.

The binomial theorem gives, for positive integers $n$,Terms containing $N^2$ or a higher matrix power vanish. After removing the common growth factor $2^n$, the remaining linear map is the shear mapping $I+(n/2)N$. It moves $(x,y)$ to $(x+ny/2,y)$. The single eigenvalue controls the exponential scale, while the nilpotent matrix adds a factor growing linearly in $n$.

In terms of polynomial interpolation, reduction modulo $(t-2{)}^2$ retains both a value and a derivative:Indeed, polynomial division gives $f(t)=f(2)+f^{\prime }(2)(t-2)+(t-2{)}^{2}h(t)$. This is a simple Hermite interpolation condition. For $f(t)=t^n$, it gives exactly the displayed matrix power. Repeated eigenvalues retain information about how a polynomial changes, not just where it vanishes.

For a length-$n$, dimension-$k$ cyclic code with generator polynomial $g$ of degree $n-k$, let the message polynomial satisfy $\mathrm{deg}M<k$. Subtracting the remainder upon polynomial division by $g$ makes the displayed word divisible by $g$. The remainder has degree less than $n-k$, so the highest $k$ coordinates retain the coefficients of $M$. This gives systematic encoding without losing any message information.