The Remainder and Factor Theorems, precalculus lesson

Dividing a polynomial by $x - 3$ and plugging $3$ into it look like two different jobs, yet the remainder of the division is exactly the value at $3$ . That one fact turns the question of whether $x - 3$ is a factor into the question of whether one value is zero, and it is the main way polynomials of degree three and higher get factored by hand.

Key terms

Evaluating a polynomial

Definition A polynomial in $x$ with real coefficients is an expression

\begin{aligned} P(x) &= a_n x^n + a_{n-1} x^{n-1} \\[0.4em] &\quad + \cdots + a_1 x + a_0, \end{aligned}

where $n$ is a nonnegative integer and $a_0, \ldots, a_n$ are real numbers. If $a_n \ne 0$ , the degree is $n$ and $a_n$ is the leading coefficient. For a number $c$ , the value $P(c)$ is the number you get by replacing every $x$ with $c$ . The same value comes out of the nested form, also called Horner’s method,

\begin{aligned} &P(c) \\[0.4em] &= \Big( \cdots \big( (a_n c + a_{n-1}) c + a_{n-2} \big) c \\[0.4em] &\quad + \cdots + a_1 \Big) c + a_0. \end{aligned}

In plain words Start with the leading coefficient, then repeat two steps, multiply by $c$ and add the next coefficient. No powers are needed, only $n$ multiplications and $n$ additions.

Example Take $P(x) = 2x^3 - 3x^2 + 4x - 5$ and $c = 2$ . Directly, $P(2) = 16 - 12 + 8 - 5 = 7$ . In nested form,

\begin{aligned} P(2) &= \big( (2 \cdot 2 - 3) \cdot 2 + 4 \big) \cdot 2 - 5 \\[0.4em] &= (1 \cdot 2 + 4) \cdot 2 - 5 \\[0.4em] &= 7. \end{aligned}

Dividing by x minus c

Definition Let $P(x)$ be a polynomial and $c$ a number. There is exactly one polynomial $Q(x)$ and one constant $r$ such that

\begin{gathered} P(x) = (x - c)\,Q(x) + r \\[0.4em] \text{for every } x. \end{gathered}

This is the division algorithm for the divisor $x - c$ . The remainder $r$ is a constant because it is $0$ or has lower degree than the divisor, which has degree $1$ . If $P$ has degree $n \ge 1$ , the quotient $Q$ has degree $n - 1$ . Synthetic division finds $Q$ and $r$ .

In plain words Dividing by $x - c$ always leaves a plain number behind, never an expression with $x$ in it.

Example Divide $P(x) = x^3 - 4x^2 + 2x + 5$ by $x - 3$ with synthetic division.

\begin{array}{r|rrrr} 3 & 1 & -4 & 2 & 5 \\ & & 3 & -3 & -3 \\ \hline & 1 & -1 & -1 & 2 \end{array}

The top row holds the coefficients of $P$ , and the leading coefficient comes straight down. Each entry of the middle row is $3$ times the bottom entry one column to its left, and each later bottom entry is the sum of the two entries above it. The bottom row gives $Q(x) = x^2 - x - 1$ and $r = 2$ , so

\begin{aligned} &x^3 - 4x^2 + 2x + 5 \\[0.4em] &\quad = (x - 3)(x^2 - x - 1) + 2. \end{aligned}

The Remainder Theorem

Definition If a polynomial $P(x)$ is divided by $x - c$ , the remainder is $P(c)$ . In other words,

P(x) = (x - c)\,Q(x) + P(c).

The proof is one line. Put $x = c$ into $P(x) = (x - c)Q(x) + r$ . The first term becomes $0 \cdot Q(c) = 0$ , which leaves $P(c) = r$ . The theorem holds word for word for complex coefficients and a complex number $c$ .

In plain words The leftover from dividing by $x - c$ is the value of the polynomial at $c$ . Watch the sign. Dividing by $x + 2$ means $c = -2$ , because $x + 2 = x - (-2)$ .

Example Find the remainder when $P(x) = x^4 - 3x^2 + 2x - 1$ is divided by $x + 2$ . Here $c = -2$ , so the remainder is

P(-2) = 16 - 12 - 4 - 1 = -1.

Long division agrees, since

\begin{aligned} &x^4 - 3x^2 + 2x - 1 \\[0.4em] &\quad = (x + 2)(x^3 - 2x^2 + x) - 1. \end{aligned}

Synthetic substitution

Definition The bottom row of synthetic division of $P(x)$ by $x - c$ lists exactly the intermediate numbers of the nested form at $c$ . Its last entry is the remainder, and by the Remainder Theorem it equals $P(c)$ . Using synthetic division to evaluate a polynomial this way is called synthetic substitution. Every power of $x$ below the degree needs its own coefficient in the top row, written as $0$ when that term is missing.

In plain words The same table that divides also evaluates. For large inputs or long polynomials it is faster and less error prone than computing each power.

Example Evaluate $P(x) = x^5 - 4x^4 + 3x^2 - 8$ at $x = 4$ . The terms in $x^3$ and $x$ are missing, so their coefficients are $0$ .

\begin{array}{r|rrrrrr} 4 & 1 & -4 & 0 & 3 & 0 & -8 \\ & & 4 & 0 & 0 & 12 & 48 \\ \hline & 1 & 0 & 0 & 3 & 12 & 40 \end{array}

So $P(4) = 40$ . The direct check is $1024 - 1024 + 48 - 8 = 40$ .

The Factor Theorem

Definition Let $P(x)$ be a polynomial and $c$ a number. We say $x - c$ is a factor of $P(x)$ if $P(x) = (x - c)\,Q(x)$ for some polynomial $Q(x)$ . The Factor Theorem states that

\begin{gathered} x - c \text{ is a factor of } P(x) \\[0.4em] \text{exactly when } P(c) = 0. \end{gathered}

It follows from the Remainder Theorem, because $P(x) = (x - c)Q(x) + P(c)$ , and the remainder is zero exactly when $P(c) = 0$ .

In plain words A zero of the polynomial and a linear factor are the same information. Find one and you have the other.

Example For $P(x) = x^3 - 7x + 6$ the values at $1$ , $2$ and $-3$ are

\begin{aligned} P(1) &= 1 - 7 + 6 = 0, \\[0.4em] P(2) &= 8 - 14 + 6 = 0, \\[0.4em] P(-3) &= -27 + 21 + 6 = 0. \end{aligned}

So $x - 1$ , $x - 2$ and $x + 3$ are all factors. Their product has degree $3$ and leading coefficient $1$ , like $P$ , and multiplying out shows that it is exactly $P$ ,

\begin{aligned} &x^3 - 7x + 6 \\[0.4em] &\quad = (x - 1)(x - 2)(x + 3). \end{aligned}

On the other hand $P(3) = 12 \ne 0$ , so $x - 3$ is not a factor.

Extracting linear factors

Definition Suppose $P(c) = 0$ . Synthetic division by $x - c$ gives a remainder of $0$ and a quotient $Q(x)$ with

P(x) = (x - c)\,Q(x),

where $Q$ has degree one less than $P$ . This quotient is often called the depressed polynomial. The zeros of $P$ are $c$ together with the zeros of $Q$ , so the search continues on $Q$ . If $Q(c) = 0$ as well, then $(x - c)^2$ divides $P$ , and the factor is repeated. More generally, for $a \ne 0$ the linear factor $ax - b$ divides $P(x)$ exactly when $P(b/a) = 0$ .

In plain words Each zero you find peels off one linear factor and lowers the degree by one. Once you are down to a quadratic, factor it or use the quadratic formula.

Example The number $3$ is a zero of $P(x) = 2x^3 - 3x^2 - 11x + 6$ , since $54 - 27 - 33 + 6 = 0$ . Divide by $x - 3$ .

\begin{array}{r|rrrr} 3 & 2 & -3 & -11 & 6 \\ & & 6 & 9 & -6 \\ \hline & 2 & 3 & -2 & 0 \end{array}

The quotient is $2x^2 + 3x - 2 = (2x - 1)(x + 2)$ , so

\begin{aligned} &2x^3 - 3x^2 - 11x + 6 \\[0.4em] &\quad = (x - 3)(2x - 1)(x + 2), \end{aligned}

with zeros $3$ , $\frac{1}{2}$ and $-2$ .

Extracting a factor also settles limits that look like $\frac{0}{0}$ . Dividing $x^3 - 7x + 6$ by $x - 1$ gives the quotient $x^2 + x - 6$ , and cancelling the common factor for $x \ne 1$ gives

\begin{aligned} &\lim_{x \to 1} \frac{x^3 - 7x + 6}{x - 1} \\[0.8em] &\quad = \lim_{x \to 1} (x^2 + x - 6) = -4. \end{aligned}

The calculus lesson on limit theorems explains why the cancellation is allowed.

Check yourself

Use the Remainder Theorem to find the remainder when $x^4 - 5x^2 + 3x + 2$ is divided by $x - 2$ .

Show the answer

The remainder is the value at $2$ , which is $16 - 20 + 6 + 2 = 4$ .

Find the number $k$ for which $x + 2$ is a factor of $x^3 + kx^2 + 5x + 6$ .

Show the answer

By the Factor Theorem the value at $-2$ must be zero. That value is $-8 + 4k - 10 + 6 = 4k - 12$ , so $k = 3$ . Indeed

\begin{aligned} &x^3 + 3x^2 + 5x + 6 \\[0.4em] &\quad = (x + 2)(x^2 + x + 3). \end{aligned}

Show that $x + 1$ is a factor of $P(x) = x^3 - 4x^2 + x + 6$ , then factor $P(x)$ completely.

Show the answer

The value $P(-1) = -1 - 4 - 1 + 6 = 0$ , so $x + 1$ is a factor. Synthetic division by $x + 1$ gives the quotient $x^2 - 5x + 6 = (x - 2)(x - 3)$ , so

\begin{aligned} &x^3 - 4x^2 + x + 6 \\[0.4em] &\quad = (x + 1)(x - 2)(x - 3). \end{aligned}

The Remainder and Factor Theorems

Key terms

Evaluating a polynomial

Dividing by x minus c

The Remainder Theorem

Synthetic substitution

The Factor Theorem

Extracting linear factors

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus