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Rational Root Theorem Christopher Boo, Brilliant Member, abc xyz, and 7 others contributed
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The rational root theorem describes a relationship between the roots of a polynomial and its coefficients. Specifically, it describes the nature of any rational roots the polynomial might possess.
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Rational Root Theorem Rational Root Theorem - Polynomial Zeros
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Contents
Algebra
Statement of the Theorem
Rational Root Theorem
Proof Integer Corollary Problem Solving
Statement of the Theorem D EF I N I T I ON
a
The rational root theorem states that if a polynomial with integer coefficients f(x) = p nxn + p n − 1x n− 1 + + p 1x + p 0 has a rational root of the form r = ± with gcd (a, b) = 1, b
then a | p 0 and b | p n.
Let's work through some examples followed by problems to try yourself.
EXAM PLE
Factorize the cubic polynomial f(x) = 2x 3 + 7x2 + 5x + 1 over the rational numbers.
a
1
1
b
1
2
By the rational root theorem, any rational root of f(x) has the form r = , where a | 1 and b | 2. Thus, we only need to try numbers ± , ± . Substituting all the possible values, f(1) > 0 f( − 1) = − 2 + 7 − 5 + 1 = 1 ≠ 0
() ( )
1 >0 2 1 2 7 5 f − = − + − + 1 = 0. 2 8 4 2 f
By the remainder-factor theorem, (2x + 1) is a factor of f(x), implying f(x) = (2x + 1)(x 2 + 3x + 1). We can then use the quadratic formula to factorize the quadratic if irrational roots are desired.
EXAM PLE
Show that √2 is irrational using the rational root theorem.
Since √2 is a root of the polynomial f(x) = x 2 − 2 , the rational root theorem states that the rational roots of f(x) are of the form ±
1, 2 1
.
None of these are roots of f(x) , and hence f(x) has no rational roots. Thus, √2 is irrational. Using this same logic, one can show that √3 , √5 , √7 , . . . are irrational, and from this one can prove that the square root of any number that is not a perfect square is irrational.
T R Y I T YOU R SELF
A polynomial with integer coefficients P(x) = a nxn + a n − 1x n− 1 + + a 0
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, with a n and a 0 being coprime positive integers, has one of the roots
2 3
. Find the second smallest possible value of a 0 + a n .
For the complete set, click here.
T R Y I T YOU R SELF
According to rational root theorem, which of the following is always in the list of possible roots of a polynomial with integer coefficients?
0 -1 4 0.5 2 0.25
T R Y I T YOU R SELF
How many rational roots does x 1000 − x 500 + x100 + x + 1 = 0
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have?
Proof PR OOF
Suppose
a b
is a root of f(x) . Then pn
() a
n
b
+ p n− 1
() a
n−1
b
+ + p1
a b
+ p 0 = 0.
By shifting the p 0
term to the right hand side, and multiplying throughout by b n , we obtain p na n + p n − 1a n− 1b + … + p 1ab n − 1 = − p 0b n . Notice that the left hand side is a multiple of a , and thus a | p 0b n . Since gcd (a, b) = 1 , Euclid's lemma implies a | p 0 . Similarly, if we shift the p n
term to the right hand side and multiply throughout by b n , we obtain p n − 1a n − 1b + p n − 2a n− 2b 2 + + p 1ab n − 1 + p 0b n = − p na n. Note that the left hand side is a multiple of b , and thus b | p na n . Since gcd (a, b) = 1 , Euclid's lemma implies b | p n . In particular, this tells us that if we want to check for 'nice' rational roots of a polynomial f(x) , we only need to check finitely many numbers of the form ±
a b
, where a | p 0 and b | p n . This is a great tool for factorizing polynomials.
Integer Corollary These are some of the associated theorems that closely follow the rational root theorem. The first one is the integer root theorem.
D EF I N I T I ON
If f(x) is a monic polynomial (leading coefficient of 1), then the rational roots of f(x) must be integers.
PR OOF
By the rational root theorem, if r =
a b
is a root of f(x) , then b | p n . But since p n = 1 by assumption, b = 1 and thus r = a is an integer.
A short example shows the usage of the integer root theorem:
EXAM PLE
Show that if x is a positive rational such that x 2 + x is an integer, then x must be an integer.
Let x 2 + x = n , where n is an integer. This is equivalent to finding the roots of f(x) = x 2 + x − n . Since f(x) is a monic polynomial, by the integer root theorem, if x is a rational root of f(x) , then it is an integer root.
Here is another theorem:
T H EOR EM
If f(x) is a polynomial with integral coefficients, a is an integral root of f(x) , and m is any integer different from a , then a − m divides f(m) .
PR OOF
On dividing f(x) by x − m, we get f(x) = (x − m)q(x) + f(m) , where q(x) is a polynomial with integral coefficients. For x = a , we get f(a) = 0 = (a − m)q(a) + f(m) or f(m) = − (a − m)q(a) . Hence a − m divides f(m) .
EXAM PLE
Let f(x) be a polynomial, having integer coefficients, and let f(0) = 1989 and f(1) = 9891 . Prove that f(x) has no integer roots.
If a is an integer root of f(x) , then a ≠ 0 as f(0) ≠ 0 . Also a must be odd since it must divide the constant term, i.e. f(0) = 1989 . But a ≠ 1 , as f(1) ≠ 0 . So taking m = 1 and using the above theorem, we see that the even number (a − 1) divides the odd number f(1) = 9891 , a contradiction. Hence f(x) has no integer roots.
Give the following problem a try to check your understandings with these theorems:
T R Y I T YOU R SELF
Find the sum of all the rational roots of the equation
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x 5 − 4x 4 + 2x3 + 2x 2 + x + 6 = 0.
Problem Solving EXAM PLE
Find all rational zeroes of P(x) = 2x 4 + x3 − 19x 2 − 9x + 9 .
Using rational root theorem, we have the following: Factors of the constant term are ± 1, ± 3, ± 9. Factors of the leading coefficient are ± 1, ± 2. Possible values of rational roots are or simply ± 1,
±1 2
, ± 3,
±3 2
, ± 9,
±9 2
±1 1
,
±1 2
±3
,
1
,
±3 2
,
±9 1
,
±9 2
,
.
Now, substituting these values in P(x) and checking if it equates to zero (please refer to this: Remainder Factor Theorem), we find that P(x) = 0 1
for the values , 3, − 3, 1. 2
Therefore, the rational zeroes of P(x) 1
are − 3, − 1, , 3. 2
T R Y I T YOU R SELF
Consider all polynomials with integral coefficients
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a nxn + a n − 1x n− 1 + + a 1x + a 0 = 0, such that
4 3
is one of its roots, 3 | a 0, and 4 | a n . Over all such polynomials, find the smallest positive value of a n + a 0 .
T R Y I T YOU R SELF
Submit your answer
12x 4 − 56x3 + 89x 2 − 56x + 12 = 0 Find the sum of real roots x that satisfy the equation above. Give your answer to 2 decimal places.
Try my other algebra problems here.
T R Y I T YOU R SELF
Given that p
Insufficient data
and q are both prime, which of the following answer choices is true about the equation px 2 − qx + q = 0 ?
Has no rational solutions Has rational solutions Has exactly one solution
T R Y I T YOU R SELF
Neither of them Brilli Not enough information Brilli the Ant is playing a game with Brian Till, her best friend. They are very competitive and always want to beat each other. Today, they are going to play the quadratic game.
Brian
Brilli is going to pick 3 non-zero real numbers and Brian is going to arrange the three numbers as the coefficients of a quadratic equation: ____ x2 + ____ x + ____ = 0. Brilli wins the game if and only if the resulting equation has two distinct rational solutions. Who has a winning strategy?
T R Y I T YOU R SELF
Submit your answer
x 4 + 3x3 + 4x 2 + 3x + 1 = 0 Let a, b, c, and d be the not necessarily distinct roots of the equation above. Find the value of the expression below: a 1024 + b 1024 + c 1024 + d 1024 +
1 a 1024
+
1 b 1024
+
1 c 1024
+
1 d 1024
.
T R Y I T YOU R SELF
A polynomial with integer coefficients P(x) = a mxm + a m − 1x m − 1 + + a 0 , with a m
n+6
and a 0 being positive integers, has one of the roots . Find the n th smallest (n
2 3
≥ 10)
n+7
possible value of a 0 + a m . Cannot be determined For the complete set, click here. n+2
n+5
n+4
n+3
Cite as: Rational Root Theorem. Brilliant.org. Retrieved 05:36, December 15, 2017, from https://brilliant.org/wiki/rational-root-theorem/
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