Difference between revisions of "Factor Theorem"

Revision as of 21:47, 13 January 2024

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 $P(a)=0$ .

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

Intermediate

Olympaid

1975 USAMO Problem 3

This article is a stub. Help us out by expanding it.

@@ Line 5: / Line 5: @@
 ==Statement==
-The '''Factor Theorem''' says that if <math>P(x)</math> is a [[polynomial]], then <math>{x-a}</math> is a [[factor]] of <math>P(x)</math> if <math>P(a)=0</math>.
+The '''Factor Theorem''' says that if <math>P(x)</math> is a [[polynomial]], then <math>x-a</math> is a [[factor]] of <math>P(x)</math> if <math>P(a)=0</math>.
 ==Proof==

Page

Toolbox

Search