Difference between revisions of "Factor Theorem"
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− | 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>. | + | In algebra, the '''Factor theorem''' is a theorem regarding the relationships between the factors of a polynomial and its roots. |
+ | |||
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+ | 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 <math>f(a)</math> (<math>a</math> is constant, <math>f</math> is polynomial) is <math>0</math> using polynomial division rather than plugging in large values. | ||
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+ | ==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 and only if <math>P(a)=0</math>. | ||
==Proof== | ==Proof== | ||
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Now suppose that <math>P(a) = 0</math>. | Now suppose that <math>P(a) = 0</math>. | ||
− | Apply [[ | + | Apply [[Remainder Theorem]] to get <math>P(x) = (x - a)Q(x) + R(x)</math>, where <math>Q(x)</math> is a polynomial with <math>\deg(Q(x)) = \deg(P(x)) - 1</math> and <math>R(x)</math> is the remainder polynomial such that <math>0\le\deg(R(x)) < \deg(x - a) = 1</math>. This means that <math>R(x)</math> can be at most a [[constant]] polynomial. |
− | Substitute <math>x = a</math> and get <math>P(a) = (a - a)Q(a) + R(a) = 0\Rightarrow R(a) = 0</math>. Since <math>R(x)</math> is a constant polynomial, <math>R(x) = 0</math> for all <math>x</math>. | + | Substitute <math>x = a</math> and get <math>P(a) = (a - a)Q(a) + R(a) = 0\Rightarrow R(a) = 0</math>. Since <math>R(x)</math> is a constant polynomial, <math>R(x) = 0</math> for all <math>x</math>. |
Therefore, <math>P(x) = (x - a)Q(x)</math>, which shows that <math>x - a</math> is a factor of <math>P(x)</math>. | Therefore, <math>P(x) = (x - a)Q(x)</math>, which shows that <math>x - a</math> is a factor of <math>P(x)</math>. | ||
− | {{ | + | ==Problems== |
+ | Here are some problems that can be solved using the Factor Theorem: | ||
+ | ===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, \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 | ||
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+ | <math>(2+r_1)(2+r_2) \cdots (2+r_{10000000010})=m!</math> | ||
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+ | for an integer <math>m</math>. If <math>p</math> is the minimum possible positive integral value of | ||
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+ | <math>(1+r_1)(1+r_2) \cdots (1+r_{10000000010})</math>. | ||
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+ | 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,\cdots,n</math>, determine <math>P(n+1)</math>. | ||
+ | |||
+ | (Source: [[1975 USAMO Problems/Problem 3|1975 USAMO Problem 3]]) | ||
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+ | ==See Also== | ||
+ | *[[Polynomials]] | ||
+ | |||
+ | *[[Remainder Theorem]] | ||
+ | |||
+ | [[Category:Algebra]] | ||
+ | |||
+ | [[Category:Polynomials]] | ||
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[[Category:Theorems]] | [[Category:Theorems]] | ||
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+ | |||
+ | {{stub}} |
Latest revision as of 12:31, 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 (
is constant,
is polynomial) is
using polynomial division rather than plugging in large values.
Statement
The Factor Theorem says that if is a polynomial, then
is a factor of
if and only if
.
Proof
If is a factor of
, then
, where
is a polynomial with
. Then
.
Now suppose that .
Apply Remainder Theorem to get , where
is a polynomial with
and
is the remainder polynomial such that
. This means that
can be at most a constant polynomial.
Substitute and get
. Since
is a constant polynomial,
for all
.
Therefore, , which shows that
is a factor of
.
Problems
Here are some problems that can be solved using the Factor Theorem:
Introductory
Intermediate
Suppose is a
-degree polynomial. The Fundamental Theorem of Algebra tells us that there are
roots, say
. Suppose all integers
ranging from
to
satisfies
. Also, suppose that
for an integer . If
is the minimum possible positive integral value of
.
Find the number of factors of the prime in
. (Source: I made it. Solution here)
Olympaid
If denotes a polynomial of degree
such that
for
, determine
.
(Source: 1975 USAMO Problem 3)
See Also
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