Factor Theorem: Difference between revisions

Revision as of 15:11, 9 November 2025

In algebra, the Factor theorem is a theorem regarding the relationships between the factors of a polynomial and its roots.

One of it's most important applications is if you are given that a polynomial have certain roots, you will know certain linear factors of the polynomial. Thus, you can test if a linear factor is a factor of a polynomial without using polynomial division and instead plugging in numbers. Conversely, you can determine whether a number in the form $f(a)$ ( $a$ is constant, $f$ is polynomial) is $0$ using polynomial division rather than plugging in large values.

Statement

The Factor Theorem says that if $P(x)$ is a polynomial, then $x-a$ is a factor of $P(x)$ if and only if $P(a)=0$ .

Proof

If $x - a$ is a factor of $P(x)$ , then $P(x) = (x - a)Q(x)$ , where $Q(x)$ is a polynomial with $\deg(Q(x)) = \deg(P(x)) - 1$ . Then $P(a) = (a - a)Q(a) = 0$ .

Now suppose that $P(a) = 0$ .

Apply Remainder Theorem to get $P(x) = (x - a)Q(x) + R(x)$ , where $Q(x)$ is a polynomial with $\deg(Q(x)) = \deg(P(x)) - 1$ and $R(x)$ is the remainder polynomial such that $0\le\deg(R(x)) < \deg(x - a) = 1$ . This means that $R(x)$ can be at most a constant polynomial.

Substitute $x = a$ and get $P(a) = (a - a)Q(a) + R(a) = 0\Rightarrow R(a) = 0$ . Since $R(x)$ is a constant polynomial, $R(x) = 0$ for all $x$ .

Therefore, $P(x) = (x - a)Q(x)$ , which shows that $x - a$ is a factor of $P(x)$ .

Problems

Here are some problems that can be solved using the Factor Theorem:

Introductory

Let $P(x)$ be the unique polynomial of minimal degree with the following properties:

$P(x)$ has a leading coefficient $1$ , $1$ is a root of $P(x)-1$ , $2$ is a root of $P(x-2)$ , $3$ is a root of $P(3x)$ , and $4$ is a root of $4P(x)$ . The roots of $P(x)$ are integers, with one exception. The root that is not an integer can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime integers. What is $m+n$ ?

$\textbf{(A) }41\qquad\textbf{(B) }43\qquad\textbf{(C) }45\qquad\textbf{(D) }47\qquad\textbf{(E) }49$

Source: 2023 AMC 10A Problems/Problem 21

Intermediate

Suppose $f(x)$ is a $10000000010$ -degree polynomial. The Fundamental Theorem of Algebra tells us that there are $10000000010$ roots, say $r_1, r_2, \cdots, r_{10000000010}$ . Suppose all integers $n$ ranging from $-1$ to $10000000008$ satisfies $f(n)=n$ . Also, suppose that

$(2+r_1)(2+r_2) \cdots (2+r_{10000000010})=m!$

for an integer $m$ . If $p$ is the minimum possible positive integral value of

$(1+r_1)(1+r_2) \cdots (1+r_{10000000010})$ .

Find the number of factors of the prime $999999937$ in $p$ . (Source: I made it. Solution here)

Olympaid

If $P(x)$ denotes a polynomial of degree $n$ such that $P(k)=\frac{k}{k+1}$ for $k=0,1,2,\cdots,n$ , determine $P(n+1)$ .

(Source: 1975 USAMO Problem 3)

@@ Line 21: / Line 21: @@
 Here are some problems that can be solved using the Factor Theorem:
 ===Introductory===
+Let <imath>P(x)</imath> be the unique polynomial of minimal degree with the following properties:
+<imath>P(x)</imath> has a leading coefficient <imath>1</imath>,
+<imath>1</imath> is a root of <imath>P(x)-1</imath>,
+<imath>2</imath> is a root of <imath>P(x-2)</imath>,
+<imath>3</imath> is a root of <imath>P(3x)</imath>, and
+<imath>4</imath> is a root of <imath>4P(x)</imath>.
+The roots of <imath>P(x)</imath> are integers, with one exception. The root that is not an integer can be written as <imath>\frac{m}{n}</imath>, where <imath>m</imath> and <imath>n</imath> are relatively prime integers. What is <imath>m+n</imath>?
+<imath>\textbf{(A) }41\qquad\textbf{(B) }43\qquad\textbf{(C) }45\qquad\textbf{(D) }47\qquad\textbf{(E) }49</imath>
+Source: [[2023 AMC 10A Problems/Problem 21]]
 ===Intermediate===

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