# Need help with Proof and conjugate roots theorem

• type_speed
In summary: But what is the something for (x+3) and (x-1)?The something is just the 0 in the equation. So (x+3) = 0 and (x-1) = 0.
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Need Assistance w/ existence of conjugate roots in polynomial

Prove:
given that f(x) = Anx^n + An-1X^n-1+ ...A1X + A0 , and An does not = 0
=> if f(x) has a root of the form (A+Bi), then it must have a root of the form (A-iB). (complex roots)

So far I've come up with the conjugate roots theorem to help me where it states that if a function f(x) is given with real coefficients then the conjugate must exist also. My professor suggested I try the Remainder Theorem??

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What exactly are you stuck on? By inspection you know that if (A+Bi) divides your polynomial then (A-Bi) must also divide your polynomial because your polynomial does not have any complex coefficients. So, I'd recommend doing either a direct or indirect proof - though it seems as if a direct proof will be an easier way to go. Assume that (A+Bi) divides your polynomial. It follows then that:

$$A_{n}x^n + A_{n-1}x^{n-1} + ... + A_{1}x + A_{0} = 0 (mod (A+Bi))$$

The rest should fall out from there...

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Appreciate the fast reply, Well first I'm stuck on why my professor told me to use the Remainder Theorem where if a polynomial f(x) is divided by x-c then f(x) is the remainder. Second, I'm not quite sure I can prove that the conjugate exists with only the given statement of the polynomial function not having any complex coefficients. Third, I doubt that our professor would assign such a easy problem for homework. (He only assigned one problem)

What does (mod(A+Bi)) mean?

I am assuming that the roots are (x-(A+Bi))= 0 and (x+(A-Bi))= 0? Right?

Tell me what you think. Regards, Chris

Mod is just a fancy way of saying remainder really. Example 4 mod 2 = 0.

Let's look at a simple polynomial as an example.

Say 0 = x^2 + 2x -3
Your solution would be (x+3)(x-1) = 0.
So your roots are x= -3, and x = 1.

That means something * (x+3) gives you x^2 + 2x -3.
And something * (x-1) gives you x^2 + 2x -3.

In this case we know what the something is, as we put in all the numbers.

## What is the Proof and Conjugate Roots Theorem?

The Proof and Conjugate Roots Theorem is a mathematical theorem that states that if a polynomial equation with real coefficients has a complex root, then its conjugate must also be a root. This theorem helps in finding all the roots of a polynomial equation.

## How is the Proof and Conjugate Roots Theorem used?

The Proof and Conjugate Roots Theorem is used in finding the roots of a polynomial equation, especially when the equation has complex roots. It also helps in simplifying the process of finding all the roots by eliminating the need for trial and error.

## Why is the Proof and Conjugate Roots Theorem important?

The Proof and Conjugate Roots Theorem is important because it helps in solving polynomial equations, which are used in various fields, including science, engineering, and economics. It also provides a general rule for finding roots, making the process more efficient and accurate.

## What are conjugate roots?

Conjugate roots are a pair of complex roots of a polynomial equation, where the imaginary parts are equal in magnitude but have opposite signs. For example, if a polynomial equation has a root of 2+3i, its conjugate root would be 2-3i.

## Can the Proof and Conjugate Roots Theorem be applied to all polynomial equations?

Yes, the Proof and Conjugate Roots Theorem can be applied to all polynomial equations with real coefficients. However, it only applies to equations with complex roots, as real roots have no conjugate.

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