Polynomial and Rational Functions - Maths: an open, verified university mathematics reference

Why this matters¶

A metal cube is heated and expands. When its side is $x$ centimetres, its volume is $x^3$ cubic centimetres. As the side grows from 2 to $x$ , the volume grows by $x^3 - 8$ , so on average the volume grows by

\frac{x^3 - 8}{x - 2}

(1)

cubic centimetres per centimetre of side. How fast does the volume grow when the side is exactly 2? Putting $x = 2$ into the formula gives $\frac{0}{0}$ , which is not a number. But the numerator factors:

x^3 - 8 = (x - 2)\,(x^2 + 2x + 4),

(2)

as multiplying out the right-hand side confirms. So for every $x \ne 2$ we may cancel the non-zero number $x - 2$ : the average rate is $x^2 + 2x + 4$ , a formula that makes sense at every $x$ , and for $x$ close to 2 it is close to $2^2 + 2 \cdot 2 + 4 = 12$ .

Why does $x^3 - 8$ have the factor $x - 2$ ? Because $2^3 - 8 = 0$ . The factor theorem of this page says that this always happens: for every polynomial $p$ and every number $a$ , the difference $p(x) - p(a)$ has the factor $x - a$ . Limits and derivatives of polynomials rest on exactly this.

The page also answers two questions that recur throughout calculus. How many solutions can an equation $p(x) = 0$ have? At most the degree of $p$ . And for which $x$ is a quotient of polynomials, such as $\frac{x^2 - 1}{x^2 - x - 6}$ , positive? Its factors decide, as soon as we know the signs of products and quotients.

Polynomials and rational functions¶

A polynomial is built from the variable $x$ and constants by adding, subtracting and multiplying, finitely many times. Collecting equal powers of $x$ brings it to a standard form.

In words. The degree is the highest power of $x$ that really occurs. Polynomials of degree 0 are the non-zero constants; those of degree 1, 2 and 3 are called linear, quadratic and cubic. A polynomial is defined at every real number: its domain is $\R$ . Any polynomial other than the zero polynomial can be written with $c_n \ne 0$ , by leaving out the terms with coefficient 0 at the top. Could one function have two such forms, with different degrees? No: Corollary 1 below shows that the values of a polynomial determine its coefficients, so its degree is well defined.

Example. $p(x) = 3x^4 - x + 7$ has degree 4, leading coefficient 3 and constant term 7. The polynomial $(x - 1)(x + 2) = x^2 + x - 2$ has degree 2, and its roots are 1 and ${-2}$ . The polynomial $x^2 - 2$ has the roots $\sqrt{2}$ and ${-\sqrt{2}}$ , because both square to 2 (the remark on square roots).

Non-example. $f(x) = \dfrac{1}{x}$ is not a polynomial. Its natural domain leaves out 0, while every polynomial is defined at 0. In the same way $\sqrt{x}$ , whose natural domain is $[0, \infty)$ , is not a polynomial.

Dividing one polynomial by another gives the next class of functions.

In words. A rational function is a quotient of two polynomials. The formula $\frac{p(x)}{q(x)}$ is a real number exactly when the denominator $q(x)$ is not 0, so the domain is $\R$ without the roots of $q$ . Every polynomial $p$ is a rational function, with $q(x) = 1$ .

Example. $f(x) = \dfrac{x + 1}{x^2 - 4}$ is a rational function. An exercise on Functions and Their Graphs asks for its natural domain; now we can also say why the denominator is 0 only at $\pm 2$ . It is $x^2 - 4 = (x - 2)(x + 2)$ , which is 0 exactly when $x = 2$ or $x = -2$ (part (a) of the sign rules below), so $\dom f = (-\infty, -2) \cup (-2, 2) \cup (2, \infty)$ .

Non-example. $g(x) = \dfrac{\sqrt{x}}{x + 1}$ is not a rational function. Its natural domain is $[0, \infty)$ , while the domain of a rational function leaves out only finitely many numbers (part (a) of the proposition on the domain below).

Main results¶

Signs of products and quotients¶

To find where a rational function is positive, we need the signs of products and quotients. The order rules, on Real Numbers and Intervals, speak of inequalities, not of signs; the proposition below derives from them the facts about signs that we use.

Proposition 1 (Sign rules)

Let $u$ and $v$ be real numbers.

(a) $uv = 0$ exactly when $u = 0$ or $v = 0$ . If $v \ne 0$ , then $\frac{u}{v} = 0$ exactly when $u = 0$ .

(b) For real numbers $x$ and $t$ , the number $x - t$ is negative if $x < t$ , zero if $x = t$ , and positive if $x > t$ .

(c) If $u \ne 0$ and $v \ne 0$ , then $uv$ and $\frac{u}{v}$ are positive when $u$ and $v$ have the same sign, and negative when their signs differ. In particular $\frac{1}{v}$ has the same sign as $v$ , $u^2 > 0$ for $u \ne 0$ , so $u^2 \ge 0$ for every real $u$ , and $1 > 0$ .

(d) A product of finitely many non-zero numbers, or a quotient of two such products, is positive if an even number of all its factors (in the numerator and the denominator together) are negative, and negative if an odd number are.

Proof

Parts (a) and (b) are short computations, (c) applies the rules for multiplying an inequality to $0 < u$ , $u < 0$ , $0 < v$ or $v < 0$ , and for (d) we multiply in the factors one at a time and follow the sign with (c).

(a) If $u = 0$ or $v = 0$ , then $uv = 0$ . Conversely, let $uv = 0$ . If $u \ne 0$ , then $v = \frac{1}{u} \cdot (uv) = \frac{1}{u} \cdot 0 = 0$ . So $u = 0$ or $v = 0$ . Now let $v \ne 0$ . Then $\frac{1}{v} \ne 0$ , because $v \cdot \frac{1}{v} = 1 \ne 0$ , and $\frac{u}{v} = u \cdot \frac{1}{v}$ . By what we have just proved, this is 0 exactly when $u = 0$ .

(b) By property 3 of the order rules, adding $-t$ to both sides keeps an inequality: $x < t$ gives $x - t < 0$ , and $x > t$ gives $x - t > 0$ . If $x = t$ , then $x - t = 0$ .

(c) First the product. Since $u \ne 0$ and $v \ne 0$ , each of them is positive or negative, by property 1 (trichotomy) of the order rules with 0 for $b$ . If $v > 0$ , multiplying $0 < v$ by $u$ gives $0 = u \cdot 0 < uv$ when $u > 0$ (property 5(a) of the order rules), and multiplying $u < 0$ by $v > 0$ gives $uv < 0 \cdot v = 0$ when $u < 0$ (property 5(a) again). If $v < 0$ , multiplying $u < 0$ by $v$ reverses it, $uv > 0 \cdot v = 0$ , when $u < 0$ (property 5(c)), and multiplying $0 < u$ by $v$ gives $0 > uv$ when $u > 0$ (property 5(c) again). So $uv > 0$ when the signs agree and $uv < 0$ when they differ.

Next, $\frac{1}{v}$ has the sign of $v$ . If $v > 0$ , then $\frac{1}{v} > 0$ (property 6 of the order rules). If $v < 0$ , adding $-v$ to both sides gives $0 < -v$ (property 3), so $\frac{1}{-v} > 0$ by property 6, and adding $\frac{1}{v}$ to both sides (property 3) gives $\frac{1}{v} < \frac{1}{-v} + \frac{1}{v} = 0$ . Since $\frac{u}{v} = u \cdot \frac{1}{v}$ and $\frac{1}{v}$ has the sign of $v$ , the rule for products gives the rule for quotients.

In particular, $u \cdot u$ has two factors of the same sign, so $u^2 > 0$ for $u \ne 0$ ; with $0^2 = 0$ this gives $u^2 \ge 0$ for every real $u$ . And $1 = 1^2 > 0$ , as $1 \ne 0$ .

(d) Start from 1, which is positive by (c), and multiply in the factors one at a time. Each partial product is non-zero, by (a). By (c), multiplying a non-zero number by a positive factor keeps its sign, and multiplying it by a negative factor changes its sign. So the sign changes once for each negative factor, and the product is positive after an even number of changes and negative after an odd number. A quotient $\frac{N}{D}$ of two such products is, by (c) again, positive exactly when $N$ and $D$ have the same sign, that is, exactly when $N \cdot D$ is positive. Since $N \cdot D$ is the product of all the factors, the count for the product decides the sign of the quotient.

In words. “Minus times minus is plus”: only the number of negative factors matters, and a quotient behaves like a product. Part (b) says that the factor $x - t$ is negative to the left of $t$ and positive to the right of it.

Degree of a product¶

The first use of the sign rules is the degree of a product of polynomials, on which the division of polynomials below rests.

Proof

We multiply out and find the highest power of $x$ that occurs.

Write $f(x) = a_m x^m + \dots + a_1 x + a_0$ with $a_m = a$ , and $g(x) = b_k x^k + \dots + b_1 x + b_0$ with $b_k = b$ . Multiplying out, $f(x)\, g(x)$ is the sum of the terms $a_i b_j\, x^{i + j}$ for $i = 0, 1, \dots, m$ and $j = 0, 1, \dots, k$ , and collecting equal powers of $x$ writes it in the form of Definition 1: it is a polynomial. Since $i \le m$ and $j \le k$ , every power that occurs has $i + j \le m + k$ , and if $i < m$ or $j < k$ , then $i + j < m + k$ (property 4 of the order rules). So only the term with $i = m$ and $j = k$ has the power $x^{m + k}$ , and its coefficient is $ab$ . Since $a \ne 0$ and $b \ne 0$ , part (a) of the sign rules gives $ab \ne 0$ . So the product has degree $m + k$ and leading coefficient $ab$ ; having a degree, it is not the zero polynomial (its form is unique by the corollary “Polynomials that agree at many points” below, whose proof does not use this proposition).

In words. Multiplying polynomials adds their degrees and multiplies their leading coefficients. For example, $(3x^2 + 1)(x - 5)$ has degree $2 + 1 = 3$ and leading coefficient $3 \cdot 1 = 3$ .

Factors and roots¶

Theorem 1 (Factor theorem)

Let $p$ be a polynomial and $a$ a real number.

(a) There is a polynomial $q$ such that, for every real $x$ ,

p(x) - p(a) = (x - a)\, q(x) ,

(6)

and so, for every real $x \ne a$ ,

\frac{p(x) - p(a)}{x - a} = q(x) .

(7)

If $p$ has degree $n \ge 1$ , then $q$ has degree $n - 1$ and the same leading coefficient as $p$ .

(b) If $p(a) = 0$ , then $p(x) = (x - a)\, q(x)$ for every real $x$ , with $q$ as in (a). Conversely, if $p(x) = (x - a)\, q(x)$ for every real $x$ , for some polynomial $q$ , then $p(a) = 0$ . So $a$ is a root of $p$ exactly when $x - a$ is a factor of $p$ .

Proof

We write $p(x) - p(a)$ as a sum of multiples of $x^k - a^k$ and take the factor $x - a$ out of each of them.

Write $p(x) = c_n x^n + \dots + c_1 x + c_0$ with $n \ge 1$ (a constant polynomial can be written so too, with $c_n = \dots = c_1 = 0$ ). The constant terms cancel in the difference:

\begin{aligned} &p(x) - p(a) \\ &\quad = c_n (x^n - a^n) + \dots \\ &\qquad + c_2 (x^2 - a^2) + c_1 (x - a). \end{aligned}

(8)

For each integer $k \ge 1$ , let

\begin{aligned} q_k(x) &= x^{k-1} + x^{k-2} a + \dots \\ &\quad + x a^{k-2} + a^{k-1}, \end{aligned}

(9)

the sum of the $k$ terms $x^{j} a^{k - 1 - j}$ for $j = 0, 1, \dots, k - 1$ . Here, as in the constant term $c_0 = c_0 x^0$ of a polynomial, $x^0$ and $a^0$ mean 1, also when $x = 0$ or $a = 0$ : we read 0⁰ as 1. For $k = 1$ , $q_1$ is the constant 1.

Then $x^k - a^k = (x - a)\, q_k(x)$ . Indeed, multiplying $q_k(x)$ by $x$ gives $x^k + x^{k-1} a + \dots + x a^{k-1}$ , multiplying it by $-a$ gives $-x^{k-1} a - \dots - x a^{k-1} - a^k$ , and in the sum of the two every term cancels except $x^k$ and $-a^k$ .

Hence $p(x) - p(a) = (x - a)\, q(x)$ for every real $x$ , with

q(x) = c_n\, q_n(x) + \dots + c_1\, q_1(x).

(10)

Since $a$ is a fixed number, each $q_k$ is a polynomial in $x$ whose highest power is $x^{k-1}$ , with coefficient 1. So $q$ is a polynomial, no power higher than $x^{n-1}$ occurs in it, and $x^{n-1}$ occurs only in $c_n\, q_n(x)$ , with coefficient $c_n$ . If $p$ has degree $n \ge 1$ , then $c_n \ne 0$ , so $q$ has degree $n - 1$ and leading coefficient $c_n$ . Finally, for $x \ne a$ the number $x - a$ is not 0, so we may divide both sides of $p(x) - p(a) = (x - a)\, q(x)$ by it, which gives the quotient form. This proves (a).

(b) If $p(a) = 0$ , the identity of (a) reads $p(x) = (x - a)\, q(x)$ . Conversely, if $p(x) = (x - a)\, q(x)$ for every real $x$ , then at $x = a$ it gives $p(a) = 0 \cdot q(a) = 0$ .

In words. Every root $a$ splits off a linear factor $x - a$ , and what is left has degree one less. The quotient $\frac{p(x) - p(a)}{x - a}$ of part (a) is undefined at $x = a$ , but it agrees with the polynomial $q$ at every other point. This is the form in which the Limits and Derivatives chapters use the theorem. The proof also gives a way to compute $q$ : replace each $x^k - a^k$ by $(x - a)\, q_k(x)$ (Example 3).

Example. For $p(x) = x^3$ and $a = 2$ , the proof gives $q(x) = q_3(x) = x^2 + 2x + 4$ , the factorisation of $x^3 - 8$ in Why this matters.

Each root splits off a factor, and the degree drops by one each time, so a polynomial cannot have more roots than its degree.

Proof

A root $r$ splits off the factor $x - r$ , and every other root is a root of the quotient, which has degree $n - 1$ . We turn this into a proof by looking at a smallest counterexample.

A polynomial of degree 0 is a non-zero constant, so it has no roots, and the theorem holds for it. Suppose the theorem fails for some degree. Then the degrees $n$ of the polynomials of degree $n$ with more than $n$ roots form a non-empty set of positive integers, and by the well-ordering principle it has a smallest element $n \ge 1$ . Let $p$ be a polynomial of degree $n$ with more than $n$ roots, and let $r$ be one of them.

By Theorem 1 (b), $p(x) = (x - r)\, q(x)$ for every real $x$ , where $q$ is the polynomial of part (a), of degree $n - 1$ . Let $s$ be a root of $p$ with $s \ne r$ . Then

0 = p(s) = (s - r)\, q(s),

(11)

and $s - r \ne 0$ , so $q(s) = 0$ by part (a) of the sign rules. So each of the roots of $p$ other than $r$ , of which there are more than $n - 1$ , is a root of $q$ . Then $q$ is a polynomial of degree $n - 1$ with more than $n - 1$ roots. If $n - 1 = 0$ , this contradicts the first paragraph; if $n - 1 \ge 1$ , it contradicts the choice of $n$ as the smallest degree with a counterexample. So the theorem holds for every degree.

The theorem says nothing about the zero polynomial, which has no degree: every real number is a root of it. For any other polynomial, it gives a test for equality.

Corollary 1 (Polynomials that agree at many points)

Let $n \in \N$ , and let

\begin{aligned} p(x) &= c_n x^n + \dots + c_1 x + c_0, \\ \tilde{p}(x) &= d_n x^n + \dots + d_1 x + d_0 \end{aligned}

(12)

be polynomials, written with the same $n$ (some coefficients may be 0). If $p(x) = \tilde{p}(x)$ at $n + 1$ different real numbers $x$ , then $c_k = d_k$ for every $k = 0, 1, \dots, n$ , and so $p(x) = \tilde{p}(x)$ for every real $x$ .

In particular, a polynomial can be written in the form of Definition 1 in only one way, apart from terms with coefficient 0: its coefficients, and its degree, are determined by its values.

Proof

The difference $p - \tilde{p}$ has $n + 1$ roots, which is too many for any polynomial of degree at most $n$ .

Let $D(x) = p(x) - \tilde{p}(x) = (c_n - d_n)x^n + \dots + (c_1 - d_1)x + (c_0 - d_0)$ . Suppose that $c_k \ne d_k$ for some $k$ , and let $m$ be the largest such $k$ . Then $D$ has degree $m$ , with leading coefficient $c_m - d_m$ , and $m \le n$ . Every number at which $p$ and $\tilde{p}$ agree is a root of $D$ , so $D$ has at least $n + 1 > m$ roots, which contradicts Theorem 2. So $c_k = d_k$ for every $k$ , and then $p$ and $\tilde{p}$ are the same formula, so they agree at every real $x$ .

For the last statement, suppose one function is given by two such formulas. Writing the shorter one with extra terms $0 \cdot x^k$ , both have the same $n$ , and they agree at every real number, in particular at $n + 1$ of them. So their coefficients are equal.

In words. Two polynomials of degree at most $n$ that agree at $n + 1$ points agree everywhere, and have the same coefficients. So we may compare coefficients: if $ax^2 + bx + c = 2x^2 - 5$ for every $x$ , then $a = 2$ , $b = 0$ and $c = -5$ .

Figure 1:Graph of $y = x^3 - 3x + c$ for $-3 \le x \le 3$ , with a slider for $c$ from ${-4}$ to 4 in steps of 0.5, starting at $c = 0$ . At $c = 0$ the graph crosses the $x$ -axis at $x = -\sqrt{3}$ , $x = 0$ and $x = \sqrt{3}$ , the roots of $x^3 - 3x = x(x^2 - 3)$ . At $c = 2$ it meets the $x$ -axis only at $x = {-2}$ and $x = 1$ , the roots of $x^3 - 3x + 2 = (x - 1)^2 (x + 2)$ ; at $x = 1$ it touches the axis without crossing it. For every $c$ the polynomial has degree 3, so it has at most three roots.

Try this: move $c$ through its whole range. For which values of $c$ on the slider does the graph meet the $x$ -axis three times, and for which only once? Can you find a $c$ for which it meets the axis four times? Theorem 2 says why not, for every real $c$ , not only those on the slider.

Division of polynomials¶

Integers can be divided with a remainder: $17 = 5 \cdot 3 + 2$ , with a remainder 2 smaller than the divisor 3. Polynomials can be divided in the same way, with “smaller” measured by the degree.

Proof (Sketch)

Long division removes the leading term of $f$ , one term at a time. If $f$ is the zero polynomial or has smaller degree than $g$ , then $q = 0$ and $r = f$ work. Otherwise let $a x^k$ be the leading term of $f$ and $b x^m$ that of $g$ , with $k \ge m$ . Subtracting $\frac{a}{b} x^{k - m} g(x)$ , whose leading term is $a x^k$ (Proposition 2), from $f(x)$ cancels the leading term of $f$ and leaves a polynomial $f_1$ that is zero or has degree less than $k$ . Repeat with $f_1$ in place of $f$ . The degree drops at every step, so after at most $k - m + 1$ steps what is left is zero or has degree less than $m$ : that is $r$ , and $q$ is the sum of the terms $\frac{a}{b} x^{k - m}$ that were subtracted. The first worked example below shows the steps.

This sketch leaves out the induction on the degree of $f$ that makes “repeat” into a proof. Uniqueness is proved in the rigorous track below, from Corollary 1.

In words. The remainder is what is left when no more multiples of $g$ can be subtracted without raising the degree. When $r$ is the zero polynomial, $g$ is a factor of $f$ : $f(x) = g(x)\, q(x)$ . Dividing by $g(x) = x - a$ , part (a) of Theorem 1 gives the quotient and the remainder at once: $p(x) = (x - a)\, q(x) + p(a)$ , and the remainder is the constant $p(a)$ (Exercise 8).

Domain, zeros and sign of a rational function¶

Proof

Parts (a) and (c) come straight from the definition of a rational function, with the bound on the number of roots and the sign rules; for (b) we choose an interval around $a$ that no other root of $q$ reaches.

(a) By Definition 2, $\dom f = \{x \in \R : q(x) \ne 0\}$ , which is $\R$ without the roots of $q$ . Since $q$ is not the zero polynomial, it has a degree $m$ , and by Theorem 2 it has at most $m$ roots.

(b) We squeeze an interval around $a$ between the roots of $q$ nearest to $a$ . By (a), the roots of $q$ other than $a$ are finitely many. If some of them are less than $a$ , let $\alpha$ be the largest of these; otherwise let $\alpha = a - 1$ . If some are greater than $a$ , let $\beta$ be the smallest of these; otherwise let $\beta = a + 1$ . Then $\alpha < a < \beta$ , so $a$ lies in $(\alpha, \beta)$ . No root of $q$ other than $a$ lies in $(\alpha, \beta)$ : a root less than $a$ is at most $\alpha$ , and a root greater than $a$ is at least $\beta$ . So every $x$ in $(\alpha, \beta)$ with $x \ne a$ has $q(x) \ne 0$ , that is, $x \in \dom f$ . If moreover $q(a) \ne 0$ , then $a \in \dom f$ as well.

(c) For $x \in \dom f$ we have $q(x) \ne 0$ , so by part (a) of the sign rules, $\frac{p(x)}{q(x)} = 0$ exactly when $p(x) = 0$ .

In words. A rational function is defined everywhere except at finitely many points, the roots of the denominator, and around each point $a$ it is defined on a whole open interval, except perhaps at $a$ itself. It is zero exactly where the numerator is zero and the denominator is not.

Its sign can be read off its factors. A polynomial such as $x^2 + 1$ is positive at every real number: $x^2 \ge 0$ and $1 > 0$ by part (c) of the sign rules, adding 1 to both sides gives $x^2 + 1 \ge 1$ (property 3 of the order rules), and so $x^2 + 1 > 0$ (property 2).

Proposition 4 (Sign of a factored rational function)

Let

f(x) = \frac{K\, N(x)\, P(x)}{D(x)\, Q(x)} ,

(14)

where:

$K \ne 0$ is a constant;
$N(x) = (x - r_1) \cdots (x - r_k)$ and $D(x) = (x - s_1) \cdots (x - s_m)$ are products of linear factors, with $k, m \in \N$ and real numbers $r_1, \dots, r_k$ , $s_1, \dots, s_m$ , not necessarily different (for $k = 0$ or $m = 0$ the empty product is read as 1);
$P$ and $Q$ are polynomials with $P(x) > 0$ and $Q(x) > 0$ for every real $x$ (either may be the constant 1).

Let $I$ be an open interval that contains none of the numbers $r_1, \dots, r_k$ , $s_1, \dots, s_m$ . Then:

$f$ is defined at every point of $I$ , and is not zero there;
$f$ has the same sign at every point of $I$ : positive if, at one point $x$ of $I$ , an even number of the numbers $K$ , $x - r_1, \dots, x - r_k$ , $x - s_1, \dots, x - s_m$ are negative, and negative if an odd number are.

Proof

On $I$ each factor keeps one sign, and then part (d) of the sign rules decides the sign of $f$ .

Let $t$ be one of the numbers $r_i$ or $s_j$ . We show that $x - t$ is non-zero and has the same sign at every $x \in I$ . By Definition 1, the open interval $I$ is one of $(\alpha, \beta)$ , $(\alpha, \infty)$ , $(-\infty, \beta)$ and $\R$ , for real $\alpha < \beta$ . It is not $\R$ , which contains $t$ .

If $I = (\alpha, \beta) = \{x \in \R : \alpha < x < \beta\}$ , then $t \notin I$ means that $\alpha < t < \beta$ fails, so $t \le \alpha$ or $t \ge \beta$ (property 7 of the order rules). If $t \le \alpha$ , every $x \in I$ satisfies $t \le \alpha < x$ , so $x > t$ by transitivity (property 2), and $x - t > 0$ by part (b) of the sign rules. If $t \ge \beta$ , every $x \in I$ satisfies $x < \beta \le t$ , so $x - t < 0$ .
If $I = (\alpha, \infty)$ , then $t \notin I$ means $t \le \alpha$ , and every $x \in I$ has $x - t > 0$ , as in the first case above. If $I = (-\infty, \beta)$ , then $t \ge \beta$ , and every $x \in I$ has $x - t < 0$ .

So at every $x \in I$ , each factor $x - r_i$ and $x - s_j$ is non-zero, with the same sign at every point of $I$ ; $P(x)$ and $Q(x)$ are positive; and $K \ne 0$ . The denominator is a product of non-zero numbers, so it is not 0, by part (a) of the sign rules (applied to one factor after another), and $x \in \dom f$ . By part (d) of the sign rules, $f(x)$ is positive if an even number of the factors are negative and negative if an odd number are; $P(x)$ and $Q(x)$ are positive, so they do not change the count. In particular $f(x) \ne 0$ . Since every factor has the same sign at every point of $I$ , the count, and with it the sign of $f$ , is the same at every point of $I$ .

In words: the sign chart. To find the sign of a rational function:

Factor the numerator and the denominator into a constant, linear factors $x - t$ , and factors that are positive everywhere, such as $x^2 + 1$ .
Mark every number $t$ of a linear factor on the number line. They cut it into open intervals, and on each of them $f$ has one sign (Proposition 4).
On each interval, count the negative factors at one point of it (or compute $f$ there).
At the marked numbers themselves, $f$ is 0 where the numerator is 0 and the denominator is not, and undefined where the denominator is 0 (Proposition 3).

Worked examples¶

Dividing and factoring¶

Example 1 (Divide

2x^3 - 3x^2 + 4x - 1

x^2 + 1

)

Goal. Find the quotient $q$ and the remainder $r$ of Theorem 3 for $f(x) = 2x^3 - 3x^2 + 4x - 1$ and $g(x) = x^2 + 1$ , of degree 2.

Cancel $2x^3$ . The leading terms give $\frac{2x^3}{x^2} = 2x$ . Subtracting $2x \cdot g(x) = 2x^3 + 2x$ from $f(x)$ leaves $-3x^2 + 2x - 1$ .
Cancel $-3x^2$ . Now $\frac{-3x^2}{x^2} = -3$ . Subtracting $-3 \cdot g(x) = -3x^2 - 3$ from $-3x^2 + 2x - 1$ leaves $2x + 2$ .
Stop. $2x + 2$ has degree 1, less than the degree 2 of $g$ . The quotient is the sum of the multipliers, $2x - 3$ , and the remainder is $2x + 2$ .

\boxed{ \begin{aligned} q(x) &= 2x - 3, \\ r(x) &= 2x + 2 \end{aligned} }

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That is, $2x^3 - 3x^2 + 4x - 1 = (x^2 + 1)(2x - 3) + (2x + 2)$ for every real $x$ .

Check. At $x = 1$ : the left-hand side is $2 - 3 + 4 - 1 = 2$ , and the right-hand side is $2 \cdot ({-1}) + 4 = 2$ . ✓ At $x = 0$ : ${-1}$ on the left and $1 \cdot ({-3}) + 2 = -1$ on the right. ✓ Multiplying out, $(x^2 + 1)(2x - 3) = 2x^3 - 3x^2 + 2x - 3$ , and adding $2x + 2$ gives $f(x)$ . ✓

Example 2 (Factor

x^3 - 2x^2 - 5x + 6

completely)

Goal. Write $p(x) = x^3 - 2x^2 - 5x + 6$ as a product of linear factors, and find all its roots.

Find one root. We try small integers (Exercise 13 shows why the divisors of the constant term 6 are the only integers worth trying): $p(1) = 1 - 2 - 5 + 6 = 0$ . So, by Theorem 1 (b), $x - 1$ is a factor.
Divide by $x - 1$ , as in Example 1. Each step cancels the leading term of what is left:
- subtracting $x^2 (x - 1)$ from $p(x)$ leaves $-x^2 - 5x + 6$ ;
- subtracting $-x\,(x - 1)$ from that leaves $-6x + 6$ ;
- subtracting $-6\,(x - 1)$ from that leaves 0.
The remainder is 0, and the quotient is $x^2 - x - 6$ : $p(x) = (x - 1)(x^2 - x - 6)$ . The quotient has degree 2 and leading coefficient 1, as part (a) of the factor theorem says.
Factor the quadratic. Two numbers with product ${-6}$ and sum ${-1}$ are ${-3}$ and 2, so $x^2 - x - 6 = (x - 3)(x + 2)$ .
The roots. By part (a) of the sign rules, applied twice, $p(x) = 0$ exactly when one of the factors is 0: at $x = 1$ , 3 or ${-2}$ . These are all the roots, as Theorem 2 also confirms: a cubic has at most three.

\boxed{p(x) = (x - 1)(x - 3)(x + 2)}

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Check. $p(3) = 27 - 18 - 15 + 6 = 0$ ✓ and $p(-2) = -8 - 8 + 10 + 6 = 0$ . ✓ The constant term of the product is $({-1})({-3})(2) = 6$ . ✓

Example 3 (The factor

x - 1

p(x) - p(1)

)

Goal. For $p(x) = 2x^3 - 5x + 1$ , find the polynomial $q$ of Theorem 1 (a) with $p(x) - p(1) = (x - 1)\, q(x)$ , and simplify $\frac{p(x) - p(1)}{x - 1}$ for $x \ne 1$ .

The difference. $p(1) = 2 - 5 + 1 = -2$ , so $p(x) - p(1) = 2x^3 - 5x + 3$ , which is $2(x^3 - 1) - 5(x - 1)$ .
Take out $x - 1$ , as in the proof of the factor theorem. Since $x^3 - 1 = (x - 1)(x^2 + x + 1)$ , the difference is $2(x - 1)(x^2 + x + 1) - 5(x - 1)$ , and taking out the common factor $x - 1$ gives $p(x) - p(1) = (x - 1)(2x^2 + 2x - 3)$ . So $q(x) = 2x^2 + 2x - 3$ : degree 2 and leading coefficient 2, as part (a) says.
Simplify. For $x \ne 1$ the number $x - 1$ is not 0, so we may divide by it: $\frac{p(x) - p(1)}{x - 1} = 2x^2 + 2x - 3$ . The polynomial on the right is also defined at $x = 1$ , where it is $2 + 2 - 3 = 1$ .

\boxed{q(x) = 2x^2 + 2x - 3}

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and $\frac{p(x) - p(1)}{x - 1} = q(x)$ for every $x \ne 1$ .

Check. At $x = 0$ : $p(0) - p(1) = 1 + 2 = 3$ , and $(0 - 1)(0 + 0 - 3) = 3$ . ✓ At $x = 2$ : $p(2) = 16 - 10 + 1 = 7$ , so $p(2) - p(1) = 9$ , and $(2 - 1)(8 + 4 - 3) = 9$ . ✓

Domain, zeros and sign¶

Example 4 (The sign of

\frac{x^2 - 1}{x^2 - x - 6}

)

Goal. For $f(x) = \dfrac{x^2 - 1}{x^2 - x - 6}$ , find the domain, the zeros, and the intervals where $f$ is positive or negative.

Factor. $x^2 - 1 = (x - 1)(x + 1)$ and $x^2 - x - 6 = (x - 3)(x + 2)$ , so
$f(x) = \frac{(x - 1)(x + 1)}{(x - 3)(x + 2)} .$
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Domain. By part (a) of the sign rules, the denominator is 0 exactly when $x = 3$ or $x = -2$ . So $\dom f = (-\infty, -2) \cup (-2, 3) \cup (3, \infty)$ .
Zeros. By Proposition 3 (c), the zeros are the roots of the numerator that lie in the domain: $x = 1$ and $x = -1$ .

Sign. The numbers ${-2}$ , ${-1}$ , 1 and 3 cut $\R$ into five open intervals, and on each of them $f$ has one sign, by Proposition 4 (here $K = 1$ and $P = Q = 1$ ). We count the negative factors, using part (b) of the sign rules: a factor $x - t$ is negative exactly to the left of $t$ .

Interval	$x + 2$	$x + 1$	$x - 1$	$x - 3$	$f(x)$
$(-\infty, -2)$	$-$	$-$	$-$	$-$	$+$
$(-2, -1)$	$+$	$-$	$-$	$-$	$-$
$(-1, 1)$	$+$	$+$	$-$	$-$	$+$
$(1, 3)$	$+$	$+$	$+$	$-$	$-$
$(3, \infty)$	$+$	$+$	$+$	$+$	$+$

Four, two and zero negative factors give $+$ ; three and one give $-$ .

Result.

$f(x) > 0$ on $(-\infty, -2)$ , on $(-1, 1)$ and on $(3, \infty)$ ;
$f(x) < 0$ on $(-2, -1)$ and on $(1, 3)$ ;
$f(x) = 0$ at $x = -1$ and $x = 1$ , and $f$ is undefined at $x = -2$ and $x = 3$ .

Check. $f(0) = \frac{-1}{-6} = \frac{1}{6} > 0$ , and 0 lies in $(-1, 1)$ . ✓ $f(2) = \frac{3}{-4} < 0$ , and 2 lies in $(1, 3)$ . ✓ $f(4) = \frac{15}{6} > 0$ . ✓

Figure 2:Graph of $y = \frac{x^2 - 1}{x^2 - x - 6}$ for $-5 \le x \le 6$ . It has gaps at $x = {-2}$ and $x = 3$ , where the function is undefined, and next to them it is steep and leaves the view. It meets the $x$ -axis at $x = {-1}$ and $x = 1$ . A table lists the values at $x = {-3}$ , ${-1.5}$ , 0, 2 and 4, one point in each interval of the sign chart. The exact values are $\frac{4}{3}$ , ${-\frac{5}{9}}$ , $\frac{1}{6}$ , ${-\frac{3}{4}}$ and $\frac{5}{2}$ : positive, negative, positive, negative, positive, as the sign chart of Example 4 predicts.

Try this: before you look at the table, predict from the factors whether the graph lies above or below the $x$ -axis on each of the five intervals. Then compare with the graph. The graph and the table show finitely many points; the sign chart, by Proposition 4, covers every point of each interval.

Example 5 (The least material for a box)

Goal. An open box (no lid) has a square base of side $x$ metres and a volume of 4 cubic metres. Show that it needs at least 12 square metres of material, and find the only box that needs exactly 12.

Model. Let $h$ be the height in metres. The volume is $x^2 h = 4$ , so $h = \frac{4}{x^2}$ . The base has area $x^2$ and each of the four sides $xh = \frac{4}{x}$ , so the area of material, in square metres, is
$S(x) = x^2 + \frac{16}{x} .$
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A box needs $x > 0$ , so the domain is $(0, \infty)$ , from the situation.
Compare with 12. Over a common denominator,
$S(x) - 12 = \frac{x^3 - 12x + 16}{x} .$
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Factor the numerator. At $x = 2$ it is $8 - 24 + 16 = 0$ , so $x - 2$ is a factor (Theorem 1 (b)). Dividing by $x - 2$ :
- subtracting $x^2 (x - 2)$ from $x^3 - 12x + 16$ leaves $2x^2 - 12x + 16$ ;
- subtracting $2x\,(x - 2)$ from that leaves $-8x + 16$ ;
- subtracting $-8\,(x - 2)$ from that leaves 0.
So $x^3 - 12x + 16 = (x - 2)(x^2 + 2x - 8)$ , and $x^2 + 2x - 8 = (x - 2)(x + 4)$ .
Sign on $(0, \infty)$ . By step 3, $S(x) - 12$ is
$\frac{(x - 2)(x - 2)(x + 4)}{x} .$
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For $x > 0$ , the factors $x$ and $x + 4$ are positive ( $x > 0 > -4$ , and part (b) of the sign rules). On $(0, 2)$ both factors $x - 2$ are negative, and on $(2, \infty)$ both are positive. So on each of these intervals an even number of factors is negative, and $S(x) - 12 > 0$ there, by Proposition 4. At $x = 2$ the numerator is 0, so $S(2) = 12$ .

\boxed{S(x) \ge 12}

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for every $x > 0$ , with $S(x) = 12$ only at $x = 2$ . So every such box needs at least 12 square metres of material, and the only box that needs exactly 12 has a base of side 2 metres and a height of $h = \frac{4}{2^2} = 1$ metre.

Check. $S(1) = 1 + 16 = 17$ , $S(2) = 4 + 8 = 12$ and $S(4) = 16 + 4 = 20$ : all at least 12. ✓ The box with $x = 2$ and $h = 1$ has volume $2^2 \cdot 1 = 4$ . ✓ Units: the area is a sum of products of two lengths in metres, so it is in square metres. ✓

Common mistakes¶

Multiplying an inequality by a factor of unknown sign

✗ Wrong: “ $\dfrac{x + 3}{x - 1} < 2$ . Multiplying by $x - 1$ gives $x + 3 < 2x - 2$ , so $x > 5$ .”

Why: multiplying by $x - 1$ keeps the inequality only where $x - 1 > 0$ . For $x < 1$ the factor is negative and reverses it (property 5(a) and (c) of the order rules). The answer $x > 5$ misses, for example, $x = 0$ , where $\frac{0 + 3}{0 - 1} = -3 < 2$ .

✓ Right: subtract 2 and use one fraction: $\frac{x + 3}{x - 1} - 2 = \frac{5 - x}{x - 1} = \frac{(-1)(x - 5)}{x - 1}$ . Its sign chart (Proposition 4, with $K = -1$ ) gives $\frac{5 - x}{x - 1} < 0$ exactly on $(-\infty, 1) \cup (5, \infty)$ .

Rigorous track¶

Why the quotient and the remainder are unique

This is the part of Theorem 3 that its sketch leaves to this section. We subtract two divisions of $f$ by $g$ from each other and compare the coefficients of the highest power of $x$ on the two sides.

Let $g$ have degree $m$ and leading coefficient $b$ , and suppose that

\begin{aligned} f(x) &= g(x)\, q_1(x) + r_1(x) \\ &= g(x)\, q_2(x) + r_2(x) \end{aligned}

(23)

for every real $x$ , where each of $r_1$ and $r_2$ is the zero polynomial or has degree less than $m$ . Then, for every real $x$ ,

\begin{aligned} &g(x)\,\bigl(q_1(x) - q_2(x)\bigr) \\ &\quad = r_2(x) - r_1(x). \end{aligned}

(24)

On the right is a polynomial in which no power $x^j$ with $j \ge m$ has a non-zero coefficient. Suppose $q_1 - q_2$ were not the zero polynomial, say of degree $k$ with leading coefficient $e \ne 0$ . Then the left-hand side would be a polynomial of degree $m + k$ , with leading coefficient $be \ne 0$ , by Proposition 2. The two sides agree at every real number, so by Corollary 1 they have the same coefficients. But the coefficient of $x^{m + k}$ , where $m + k \ge m$ , is $be \ne 0$ on the left and 0 on the right. So $q_1 - q_2$ is the zero polynomial: $q_1(x) = q_2(x)$ for every $x$ , and then $r_1(x) = r_2(x)$ for every $x$ as well.

The degree of the zero polynomial

The zero polynomial has no highest power with a non-zero coefficient, so it has no degree. Some books give it the degree ${-\infty}$ , with the convention $-\infty + n = -\infty$ , so that “the degree of a product is the sum of the degrees” holds for every pair of polynomials. We avoid the convention and state the zero polynomial separately instead. It is the one polynomial that Theorem 2 must leave out: every real number is a root of it.

Summary¶

A polynomial $c_n x^n + \dots + c_1 x + c_0$ with $c_n \ne 0$ has degree $n$ ; its values determine its coefficients (Corollary 1). The degree of a product is the sum of the degrees (Proposition 2). A rational function $\frac{p(x)}{q(x)}$ is defined exactly where $q(x) \ne 0$ .
Factor theorem (Theorem 1): $p(x) - p(a) = (x - a)\, q(x)$ for a polynomial $q$ of degree one less, so
$\frac{p(x) - p(a)}{x - a} = q(x) \quad (x \ne a),$
(25)
and $a$ is a root of $p$ exactly when $x - a$ is a factor.
A polynomial of degree $n$ has at most $n$ roots (Theorem 2).
Division: $f = gq + r$ with $r$ zero or of smaller degree than $g$ (Theorem 3); subtract multiples of $g$ to cancel leading terms.
Signs: only the number of negative factors matters (Proposition 1). A factored rational function has one sign on each interval between the numbers of its linear factors (Proposition 4). Never multiply an inequality by a factor of unknown sign.

Exercises¶

Exercise 13 (Integer roots divide the constant term)

Let $p(x) = c_n x^n + \dots + c_1 x + c_0$ be a polynomial whose coefficients are integers, and let $r$ be an integer with $p(r) = 0$ . Show that $r$ divides $c_0$ : there is an integer $k$ with $c_0 = rk$ . Use this to find all the integer roots of $x^3 + x^2 - 7x + 2$ .

Hint 1

Move $c_0$ to one side of $p(r) = 0$ and take out a factor $r$ from the other side.

Hint 2

The integers that divide 2 are $\pm 1$ and $\pm 2$ . Test each of them.

Answer

$c_0 = r \cdot \bigl(-(c_n r^{n-1} + \dots + c_1)\bigr)$ , and the bracket is an integer. The only integer root of $x^3 + x^2 - 7x + 2$ is 2.

Where this leads¶

These topics build on this one:

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