How can you prove whether the roots of a polynomial equation (i.e. p(x) ) are irrational?

Question

for example x^5 -5x^2 + 2x - 1

Sorry I meant p(x)=0

Hmm, I'm looking for an actual *proof* that the roots of an equation are irrational, a thorough one, like the proof that the square root of two is irrational:
So possibly a proof by contradiction:
eg if we assume that root2 is rational, then it can be written as a fraction in lowest terms, i.e.
a/b = root2
so a^2/b^2 = 2
so a^2 = 2b^2
So a^2 is even, so a must be even and can be written as 2k.
so 2k/b = root2
so 4k^2/b^2 = 2
so 4k^2 = 2b^2
so b^2 must be even, so b must also b even.

so a/b is not in lowest terms, even though it was defined to be. Therefore, root2 is not rational as it cannot be written as a fraction.

why shouldn't q divide a(n) or p divide a(0)?

Answer 1

The rational roots of any polynomial:

a(n)x^n + a(n-1)x^(n-1) + ... + a(1)x + a(0)

are in the form ±p/q, where p divides a(0) and q divides a(n).
In your case, a(0) = -1 and a(n) = 1, so the only rational roots possible are ±1.
Substitute ±1 into the equation to see if either is a root. (neither is)

Example:
x^3 + 4x^2 + 2x - 4 = 0
The only possible rational roots are ±1, ±2, ±4 (These numbers divide a(0), and only 1 divides a(3)=1 ). Substituting each of these into the equation (synthetic substitution works well), we find -2 the only rational root.

*****************************************
Proof by contradiction.
Let the rational number p/q be a reduced root (i.e, q and p are relatively prime integers) to the rational polynomial equation:
a(n)x^n + a(n-1)x^(n-1) + ... + a(1)x + a(0) = 0
and such that either p does not divide a(0) or q does not divide a(n). [Note: without loss of generality we can assume all coefficients are integers by multiplying both sides by the least common multiple of the denominators of all the a(n)'s]

Then we know that
a(n)p^n/q^n + a(n-1)p^(n-1)/q^(n-1) + ... + a(1)p/q + a(0) = 0
multiply both sides by q^n to get:

a(n)p^n + a(n-1) qp(n-1) + a(n-2) q^2p^(n-2) + ... + a(1) q^(n-1)p
+ a(0)q^n = 0

or a(n)p^n + q * f(a,p,q) = 0 for some function f of a,p,q.
and a(0)q^n + p * g(a,p,q) = 0 for some function g of a,p,q.

This means
a(n)p^n = -q * f(a,p,q)
a(0)q^n = -p * g(a,p,q)

By the first equation, q must divide a(n) * p^n.
But q has no factor in common with p, so it cannot have a factor in common with p^n, so all the factors of q (ie, q itself) must divide a(n), and we assumed that q does not divide a(n), so p must divide a(0) * q^n (the only possibility left)
But likewise with the second equation, p must divide a(0) * q^n. But p does not have any common factors with q, so it cannot have common factors with q^n, so p must divide a(0) and we assumed that p does not divide a(0), so a contradiction arises.
Therefore no such p/q can exist, therefore the roots are irrational.

Q:
why shouldn't q divide a(n) or p divide a(0)?

Because otherwise if both did then the roots COULD be rational (p/q could be a root). If thats the case you prove/disprove it by example. In other words, its easy to show that in your example the roots are all irrational because the only POSSIBLE rational ones are 1 and -1. So you just substitute them in to see that they're not roots. My proof just shows that IF there are rational roots, they must be p/q where p|a(0) and q|a(n).

There is no general way to determine by inspection that a random polynomial has all irrational roots. The only way is to check the POSSIBLE rational ones by substitution.

Answer 2

Your assumption, "if 2+i is a root, then 2-i is likewise a root," is authentic assuming that the polynomial equation has authentic huge style coefficients, which it probable does given the category that you're in. once you're searching to aspect this polynomial, then you could do man made branch with one root and then with the different root. although, this may get nasty with complicated mathematics, that you'll or would no longer recognize. yet differently to handle that's to divide by the "minimum polynomial" of two+i and a couple of-i. specifically, those adult men fulfill the polynomial: [x-(2+i)][x-(2-i)] = x^2 - 4x + 5 This polynomial has to divide the unique polynomial. often, if a+bi is a root of a polynomial with authentic coefficients, then a-bi is, too, and the polynomial is divisible by: [x-(a+bi)][x-(a-bi)] = x^2 - 2ax + a^2 + b^2

Answer 3

There's several different ways.
1. Factor it. Find the roots.

2A. Differentiate it.
2B. Find the roots of this equation.
2C. If there are not 4 real roots to this equation, then there cannot be 5 roots to the previous equation. Why?
In order for us to have 5 roots for the original equation, we must cross the axis 5 times
The slope of the graph must change from negative to positive at least 4 times.
2D. Find the values of p at the roots of p', and see if it's possible for p to have 5 roots.


Artificially constructed example:
p(x) = x^5 + 16.25x^4 + 10x^3 -530x^2 + 1400x -244
A. p'(x) = 5x^4 + 65x^3 + 30x^2 -1060x + 1400

B. solving for p'(x) = 0
0 = 5x^4 + 65x^3 + 30x^2 -1060x + 1400
= x^4 + 13x^3 + 6x^2 - 212 x + 280
= (X-2)^2(x+7)(x+10)

C. 4 real roots, 2, 2, -7, -10, Continue

D. p(-10) = -14744local max
p(-7) = -17234.8local min
p(2) = 808local max from left, local min from right (inflection pt)
There can be one root between -7 and 2.
p has 1 real root. thus 4 irrational.

Notice if we changed the orignal equation to
p(x) = x^5 + 16.25x^4 + 10x^3 -530x^2 + 1400x +15000
we get:
p(-10) = 500local max
p(-7) = -1990.75 local min
p(2) = 16052local max from left, local min from right (inflection pt)
Now, there can be one root before -10, one root between -10 and -7, and one between -7 and 2
p has 3 real roots. thus 2 irrational