Factor Theorem

Revision as of 12:31, 2 May 2024 by Taiko (talk | contribs) (Olympaid)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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.


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


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 $P(x)$.


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



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)


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)

See Also

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