Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Polynomial Remainder Theorem"

m (This article is still TERRIBLE but the proof makes a little more sense.)
m (typo fixes)
Line 18: Line 18:
  
 
=== More examples ===
 
=== More examples ===
−
* [[1950 AHSME Problems/Problem 20 | 1950 ASHME Problem 20]]
+
* [[1950 AHSME Problems/Problem 20 | 1950 AHSME Problem 20]]
−
* [[1961 AHSME Problems/Problem 22 | 1961 ASHME Problem 22]]
+
* [[1961 AHSME Problems/Problem 22 | 1961 AHSME Problem 22]]
−
* [[1969 AHSME Problems/Problem 34 | 1969 ASHME Problem 34]]
+
* [[1969 AHSME Problems/Problem 34 | 1969 AHSME Problem 34]]
  
 
== See also ==
 
== See also ==

Revision as of 12:18, 11 March 2025

In algebra, the Polynomial Remainder Theorem states that the remainder upon dividing any polynomial $P(x)$ by a linear polynomial $x-a$, both with complex coefficients, is equal to $P(a)$.

Proof

By polynomial division with dividend $P(x)$ and divisor $x-a$, that exist a quotient $Q(x)$ and remainder $R(x)$ such that \[P(x) = (x-a) Q(x) + R(x)\] with $\deg R(x) < \deg (x-a)$. We wish to show that $R(x)$ is equal to the constant $P(a)$. Because $\deg (x-a) = 1$, $\deg R(x) < 1$. Therefore, $\deg R(x) = 0$, and so the $R(x)$ is a constant.

Let this constant be $r$. We may substitute this into our original equation and rearrange to yield \[r = P(x) - (x-a) Q(x).\] When $x = a$, this equation becomes $r = P(a)$. Hence, the remainder upon diving $P(x)$ by $x-a$ is equal to $P(a)$. $\square$

Generalization

The strategy used in the above proof can be generalized to divisors with degree greater than $1$. A more general method, with any dividend $P(x)$ and divisor $D(x)$, is to write $R(x) = D(x) Q(x) - P(x)$, and then substitute the zeroes of $D(x)$ to eliminate $Q(x)$ and find values of $R(x)$. Example 2 showcases this strategy.

Examples

Here are some problems with solutions that utilize the Polynomial Remainder Theorem and its generalization.

Example 1

What is the remainder when $x^2+2x+3$ is divided by $x+1$?

Solution: Although one could use long or synthetic division, the Polynomial Remainder Theorem provides a significantly shorter solution. Note that $P(x) = x^2 + 2x + 3$, and $x-a = x+1$. A common mistake is to forget to flip the negative sign and assume $a = 1$, but simplifying the linear equation yields $a = -1$. Thus, the answer is $P(-1)$, or $(-1)^2 + 2(-1) + 3$, which is equal to $2$. $\square$.

More examples

See also

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Polynomial_Remainder_Theorem&oldid=244750"