Difference between revisions of "Factor Theorem"

Revision as of 11:31, 2 May 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 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

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 22: / Line 22: @@
 ===Introductory===
 ===Intermediate===
-Suppose <math>f(x)</math> is a <math>10000000010</math>-degree polynomial. The Fundamental Theorem of Algebra tells us that there are <math>10000000010</math> roots, say <math>r_1, r_2, \dots, r_{10000000010}</math>. Suppose all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)=n</math>. Also, suppose that
+Suppose <math>f(x)</math> is a <math>10000000010</math>-degree polynomial. The Fundamental Theorem of Algebra tells us that there are <math>10000000010</math> roots, say <math>r_1, r_2, \cdots, r_{10000000010}</math>. Suppose all integers <math>n</math> ranging from <math>-1</math> to <math>10000000008</math> satisfies <math>f(n)=n</math>. Also, suppose that
-<math>(2+r_1)(2+r_2) \dots (2+r_{10000000010})=m!</math>
+<math>(2+r_1)(2+r_2) \cdots (2+r_{10000000010})=m!</math>
 for an integer <math>m</math>. If <math>p</math> is the minimum possible positive integral value of
-<math>(1+r_1)(1+r_2) \dots (1+r_{10000000010})</math>.
+<math>(1+r_1)(1+r_2) \cdots (1+r_{10000000010})</math>.
 Find the number of factors of the prime <math>999999937</math> in <math>p</math>. (Source: I made it. Solution [[User:Ddk001#Problem_7|here]])
 ===Olympaid===
-If <math>P(x)</math> denotes a polynomial of degree <math>n</math> such that<cmath>P(k)=\frac{k}{k+1}</cmath>for <math>k=0,1,2,\ldots,n</math>, determine <math>P(n+1)</math>.
+If <math>P(x)</math> denotes a polynomial of degree <math>n</math> such that<cmath>P(k)=\frac{k}{k+1}</cmath>for <math>k=0,1,2,\cdots,n</math>, determine <math>P(n+1)</math>.
 (Source: [[1975 USAMO Problems/Problem 3|1975 USAMO Problem 3]])
 ==See Also==

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