Factorizing Polynomials with Irrational Exponents

[cdummie]

I should factorize following polynomial:

P(x)=x^2n + 2cos(naπ)x^n + 1 in ℝ if i know that a is irrational number.

Things that confuse me here are following:

1. When factorizing polynomials, i have known exponents (unlike here, where i have 2n and n) so i don't know what to do with them?

2. Why does it makes a difference if a is irrational number?

 

[jedishrfu]

Have you considered a substitution of variable x to make the equation a simple quadratic?

 

[cdummie]

Have you considered a substitution of variable x to make the equation a simple quadratic?

You mean like: y=xⁿ, but what can i do next, that cosine there is confusing, and fact that a is irrational makes it even more complicated.

 

[SteamKing]

You mean like: y=xⁿ, but what can i do next, that cosine there is confusing, and fact that a is irrational makes it even more complicated.

The value of the cosine in your polynomial doesn't depend on x, so it may be treated like a constant. Call it A and re-write your original polynomial:

A = cos(naπ)

P(x) = x²ⁿ + 2Axⁿ + 1

Now, make the substitution y = xⁿ

 

[cdummie]

The value of the cosine in your polynomial doesn't depend on x, so it may be treated like a constant. Call it A and re-write your original polynomial:

A = cos(naπ)

P(x) = x²ⁿ + 2Axⁿ + 1

Now, make the substitution y = xⁿ

Ok, so now i have the following:

y² + 2Ay + 1

y_1,2=[-2A+-sqrt(4A² - 4)]/2 = -A +- sqrt(A² - 1)

since A=cos(naπ) then A²=cos²(naπ) and A² - 1 = cos²(naπ) - 1 =-sin²(naπ)

but -sin²(naπ) could be anywhere between -1 and 1 so i don't know is it positive or negative, because if it is negative then i can't factorize it in ℝ.

 

[SteamKing]

Ok, so now i have the following:

y² + 2Ay + 1

y_1,2=[-2A+-sqrt(4A² - 4)]/2 = -A +- sqrt(A² - 1)

since A=cos(naπ) then A²=cos²(naπ) and A² - 1 = cos²(naπ) - 1 =-sin²(naπ)

but -sin²(naπ) could be anywhere between -1 and 1 so i don't know is it positive or negative, because if it is negative then i can't factorize it in ℝ.

Obviously, in order to factor this polynomial in ℝ, certain conditions must be met; IOW, A² - 1 ≥ 0, or A² ≥ 1 or cos²(naπ) ≥ 1.

Since -1 ≤ cos (θ) ≤ 1, then the polynomial can be factored in ℝ only if naπ = 2kπ, where k = 0, 1, 2, ...

It's not clear why you would want to factor such a beast anyway. If you want to find the roots to P(x), it would seem that a numerical approach would work just as well.

 

[cdummie]

Obviously, in order to factor this polynomial in ℝ, certain conditions must be met; IOW, A² - 1 ≥ 0, or A² ≥ 1 or cos²(naπ) ≥ 1.

Since -1 ≤ cos (θ) ≤ 1, then the polynomial can be factored in ℝ only if naπ = 2kπ, where k = 0, 1, 2, ...

It's not clear why you would want to factor such a beast anyway. If you want to find the roots to P(x), it would seem that a numerical approach would work just as well.

Well, it's just an exam question, anyway, what's with a, i mean why is it pointed out that a has to be irrational number, the way we solved this, it wouldn't be any different even if a wasn't irrational number?

 

[SteamKing]

Well, it's just an exam question, anyway, what's with a, i mean why is it pointed out that a has to be irrational number, the way we solved this, it wouldn't be any different even if a wasn't irrational number?

It's not clear what n is here. Is n any integer, or is it something else?

Since a is irrational, this suggests that the condition naπ = 2kπ can be satisfied only if na = 0, and this will depend on what values n can take.

 

[cdummie]

It's not clear what n is here. Is n any integer, or is it something else?

Since a is irrational, this suggests that the condition naπ = 2kπ can be satisfied only if na = 0, and this will depend on what values n can take.

I don't know about n, it's not defined what n is, but it might be integer (probably it is), anyway, since it is not defined i can't be sure what it is.