Factor Theorem

Revision as of 20:47, 13 January 2024 by Ddk001 (talk | contribs) (Statement)

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


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