The Remainder Theorem connects division and evaluation: the remainder when a polynomial $f(x)$ is divided by $x - b$ is exactly $f(b)$. This means synthetic division does double duty as a fast method for evaluating polynomials. Its immediate consequence, the Factor Theorem, characterizes roots in terms of factors and is the foundation for all polynomial factoring.
Quick Reference
| Result | Statement |
|---|---|
| Remainder Theorem | Remainder of $f(x) \div (x - b)$ equals $f(b)$ |
| Factor Theorem | $f(b) = 0 \iff (x-b)$ is a factor of $f(x)$ |
| Root test | To check if $b$ is a root of $f$, divide synthetically and check if $R = 0$ |
| Polynomial evaluation | $f(b)$ can be computed as the remainder of synthetic division by $x - b$ |
| Corollary | If $f(a)=0$ and $f(b)=0$, then $(x-a)(x-b)$ divides $f(x)$ |
| Degree bound | A degree-$n$ polynomial has at most $n$ roots |
The Remainder Theorem
When a polynomial $f(x)$ is divided by $x - b$, the remainder equals $f(b)$.
Proof
Let $\phi(x)$ be the quotient and $R$ the remainder when $f(x)$ is divided by $x - b$. By the division algorithm: $f(x) = \phi(x)(x - b) + R.$ Since $R$ has degree less than 1 (the degree of $x - b$), $R$ is a constant. This identity holds for all values of $x$. Setting $x = b$: $f(b) = \phi(b)(b - b) + R = \phi(b) \cdot 0 + R = R.$ Therefore $f(b) = R$.
Evaluating Polynomials via Synthetic Division
The Remainder Theorem makes synthetic division a rapid evaluation tool. To find $f(b)$, divide $f(x)$ by $x - b$ synthetically; the last entry in the bottom row is $f(b)$.
Find $f(4)$ where $f(x) = 5x^4 - 12x^3 - 20x^2 - 43x + 6$.
Solution
Direct substitution: $f(4) = 5(256) - 12(64) - 20(16) - 43(4) + 6 = 1280 - 768 - 320 - 172 + 6 = 26.$ Synthetic division by $x - 4$: $ \begin{array}{c|rrrrr} 4 & 5 & -12 & -20 & -43 & 6 \\ + & & 20 & 32 & 48 & 20 \\ \hline & 5 & 8 & 12 & 5 & 26 \end{array} $ The remainder is $26 = f(4)$. Both methods agree, confirming the Remainder Theorem.
The Factor Theorem
(Factor Theorem) A polynomial $f(x)$ vanishes when $x = b$ (i.e., $f(b) = 0$) if and only if $(x - b)$ is a factor of $f(x)$.
Proof
By the Remainder Theorem, when $f(x)$ is divided by $x - b$: $f(x) = \phi(x)(x - b) + f(b).$ If $f(b) = 0$, then $f(x) = \phi(x)(x - b)$, so $(x - b)$ is a factor. Conversely, if $(x - b)$ is a factor, the division is exact and the remainder is 0, so $f(b) = 0$.If $x^3 + 3x^2 - m$ is exactly divisible by $x - 2$, find $m$.
Solution
By the Factor Theorem, $x = 2$ must be a root: $f(2) = 8 + 12 - m = 0 \implies m = 20.$Show that $x^n + b^n$ is exactly divisible by $x + b$ if $n$ is odd, but not if $n$ is even.
Solution
By the Factor Theorem, $x + b = x - (-b)$ is a factor if and only if $f(-b) = 0$ where $f(x) = x^n + b^n$. $f(-b) = (-b)^n + b^n.$ If $n$ is odd: $(-b)^n = -b^n$, so $f(-b) = -b^n + b^n = 0$. Thus $(x+b)$ is a factor. If $n$ is even: $(-b)^n = b^n$, so $f(-b) = b^n + b^n = 2b^n$. This is zero only when $b = 0$. For $b \neq 0$, $(x + b)$ is not a factor.
Consequences of the Factor Theorem
If $f(x)$ vanishes at $x = a$ and also at $x = b$ (with $a \neq b$), then $(x - a)(x - b)$ divides $f(x)$.
Proof
Since $f(a) = 0$, $(x - a)$ is a factor: $f(x) = (x - a)\phi_1(x)$. Substituting $x = b$: $0 = f(b) = (b - a)\phi_1(b)$. Since $b \neq a$, we have $b - a \neq 0$, so $\phi_1(b) = 0$. By the Factor Theorem applied to $\phi_1$, $(x - b)$ is a factor of $\phi_1(x)$: $\phi_1(x) = (x - b)\phi_2(x)$. Therefore $f(x) = (x - a)(x - b)\phi_2(x)$.The same reasoning extends to any finite set of distinct roots:
If $f(x)$ vanishes at $x = a_1, a_2, \ldots, a_k$ (all distinct), then $(x - a_1)(x - a_2)\cdots(x - a_k)$ divides $f(x)$.
Find a degree-2 polynomial $f(x)$ that equals 0 when $x = 2$ and $x = 3$, and equals 6 when $x = 4$.
Solution
Since $f$ vanishes at $x = 2$ and $x = 3$, the factor theorem gives $f(x) = a_0(x-2)(x-3)$ for some constant $a_0$. Using $f(4) = 6$: $6 = a_0(4-2)(4-3) = 2a_0$, so $a_0 = 3$. Therefore $f(x) = 3(x-2)(x-3) = 3x^2 - 15x + 18$.Find a degree-3 polynomial $f(x)$ that vanishes at $x = 2$ and $x = 3$, equals 6 when $x = 1$, and equals 18 when $x = 4$.
Solution
Since $f$ vanishes at $x = 2$ and $x = 3$, we have $f(x) = (a_1 x + a_0)(x - 2)(x - 3)$. Using the conditions $f(1) = 6$ and $f(4) = 18$: $6 = (a_1 + a_0)(1 - 2)(1 - 3) = 2(a_1 + a_0) \implies a_1 + a_0 = 3.$ $18 = (4a_1 + a_0)(4 - 2)(4 - 3) = 2(4a_1 + a_0) \implies 4a_1 + a_0 = 9.$ Subtracting: $3a_1 = 6$, so $a_1 = 2$ and $a_0 = 1$. Therefore $f(x) = (2x + 1)(x - 2)(x - 3) = 2x^3 - 9x^2 + 7x + 6$.
Degree Bound on the Number of Roots
A polynomial $f(x)$ of degree $n$ can have at most $n$ distinct roots.
Proof
If $f$ had more than $n$ distinct roots, it would be divisible by the product of more than $n$ linear factors $(x - a_i)$. That product has degree greater than $n$, which is impossible for a degree-$n$ polynomial.An important corollary gives a very useful identity principle:
If two polynomials of degree at most $n$ agree at more than $n$ values of $x$, then they are identical: all their corresponding coefficients are equal.
This means that if you know a polynomial of degree $\leq n$ by its values at $n + 1$ or more points, you know it completely.
