notes / Integrated Mathematics / chapter 10: Polynomials
10.9 The Rational Root Theorem
The same idea extends to rational roots: a root in lowest terms has dividing the constant term and dividing the leading coefficient.
Problem 10.9.1.
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Is there with ? Possible: .
The negative candidates don’t work: , , , .
So that leaves : , ; for the first 3 terms sum to something that ends in 5, so no; .
Now suppose :
So , and
Since , . And with , so .
Similar logic for , using , whose terms all have a .
Theorem 10.9.2 (Rational Root Theorem). Given , if and are relatively prime integers and , then and .
Proof. Let be a polynomial with and . Let such that are relatively prime and . Then
Since and , . Similarly for :
∎
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Divisors: : and : , so
are the possibilities (numerator and denominator must be relatively prime).
the roots are , , .
Problem 10.9.3. Find the roots:
(a)
(b)
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(a) and , so the non-integer candidates are
Signs of the four terms:
NOT .
There is a root between and , since the sign changes.
So is a root:
Try on : , then ; even? ( isn’t an integer, so it can’t cancel the .)
Candidates let’s work backwards:
So:
(b) , over denominators giving
: and , so . .
.
(by calculator)
Candidates now: over .
and , .
are candidates, and . Yay!
With :
Check the constant term: . ✓
Roots: , , , .
Tips:
- Factor, factor, factor!
- Start your rational root hunt with .
- Try the same strats in different ways.
- Example general for proofs.
Exercises
Exercise 10.9.4. Show solution
Because for all with , there exists such that in () are relatively prime.
Exercise 10.9.5. Find consecutive integers such that the cubes of the first three add up to the cube of the last: for .
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Expanding,
which add up to , while . So
Candidates for rational roots: .
✓. Synthetic division by :
so the quotient is , whose roots are
So , and indeed :