Decomposing Polynomial P: Roots, Multiplicity, And Irreducibility

Hey guys! Let's dive into the fascinating world of polynomials! We're going to break down a specific polynomial, explore its roots, figure out their multiplicity, and finally, decompose it into irreducible factors within both complex and real number systems. Buckle up, it's going to be a mathematical adventure!

Unveiling the Polynomial P

Our star today is the polynomial P(X) = X⁵ - X⁴ + 2X³ - 2X² + X - 1. This might look intimidating at first glance, but don't worry, we'll tackle it step by step. The main goal here is to understand its structure by finding its roots and then expressing it as a product of simpler polynomials that cannot be factored further. This process, known as decomposition into irreducible polynomials, is super important in various areas of mathematics, including algebra and calculus. When we're dealing with complex numbers, our goal is to write the polynomial as a product of linear factors, each corresponding to a root. Over real numbers, we might have linear factors for real roots and quadratic factors for pairs of complex conjugate roots. Breaking it down like this gives us a much clearer picture of the polynomial's behavior and properties. So, let's roll up our sleeves and get to work!

1. Proving 'i' is a Root and Finding Its Multiplicity

First up, we need to show that the imaginary unit, i, is a root of our polynomial P(X). Remember, a root of a polynomial is a value that makes the polynomial equal to zero. So, we'll substitute i into P(X) and see what happens. Let's calculate P(i):

P(i) = (i)⁵ - (i)⁴ + 2(i)³ - 2(i)² + i - 1

Now, let's simplify this using the powers of i: i² = -1, i³ = -i, i⁴ = 1, and i⁵ = i.

So we get:

P(i) = i - 1 - 2i + 2 + i - 1 = (i - 2i + i) + (-1 + 2 - 1) = 0

Awesome! P(i) equals zero, which confirms that i is indeed a root of P(X). But we're not done yet. We need to find the multiplicity of this root. Multiplicity tells us how many times a root appears as a factor in the polynomial. To find this, we can repeatedly divide P(X) by (X - i) until it's no longer a factor. However, there's a neat trick we can use: if i is a root with multiplicity m, then it will also be a root of the first m-1 derivatives of P(X).

Let's find the first derivative, P'(X):

P'(X) = 5X⁴ - 4X³ + 6X² - 4X + 1

Now, let's plug in i into P'(X):

P'(i) = 5(i)⁴ - 4(i)³ + 6(i)² - 4i + 1 = 5 + 4i - 6 - 4i + 1 = 0

Great! P'(i) is also zero, meaning the multiplicity of i is at least 2. Now, let's find the second derivative, P''(X):

P''(X) = 20X³ - 12X² + 12X - 4

And plug in i:

P''(i) = 20(i)³ - 12(i)² + 12i - 4 = -20i + 12 + 12i - 4 = 8 - 8i ≠ 0

Since P''(i) is not zero, the multiplicity of i is exactly 2. This means (X - i)² is a factor of P(X).

2. Showing 1 is a Real Root

Next, we need to demonstrate that 1 is a real root of P(X). This is pretty straightforward – we simply substitute X = 1 into the polynomial:

P(1) = (1)⁵ - (1)⁴ + 2(1)³ - 2(1)² + 1 - 1 = 1 - 1 + 2 - 2 + 1 - 1 = 0

Bingo! P(1) = 0, confirming that 1 is indeed a real root of P(X). Now we know that (X - 1) is also a factor of P(X). Since we've already established that i is a root with multiplicity 2, its complex conjugate, -i, must also be a root with multiplicity 2. This is because polynomials with real coefficients have complex roots that come in conjugate pairs. So, (-i) is also a root with multiplicity 2.

3. Decomposing P into Irreducible Polynomials

Now for the grand finale: decomposing P(X) into irreducible polynomials, first in C[X] (the set of polynomials with complex coefficients) and then in R[X] (the set of polynomials with real coefficients).

Decomposition in C[X]

We know the roots of P(X) are 1, i (with multiplicity 2), and -i (with multiplicity 2). Therefore, we can write P(X) as a product of linear factors corresponding to these roots:

P(X) = (X - 1)(X - i)²(X + i)²

This is the complete decomposition of P(X) into irreducible polynomials in C[X]. Each factor is linear, and linear polynomials are always irreducible (they can't be factored further).

Decomposition in R[X]

For R[X], we need to combine the complex conjugate pairs into quadratic factors with real coefficients. We already have the real root 1, so (X - 1) will be one factor. Now, let's combine the factors corresponding to i and -i:

(X - i)(X + i) = X² - (i)² = X² + 1

Since we have these roots with multiplicity 2, we actually have two factors of (X² + 1).

Therefore, the decomposition of P(X) into irreducible polynomials in R[X] is:

P(X) = (X - 1)(X² + 1)²

Here, (X - 1) is a linear factor, and (X² + 1) is an irreducible quadratic factor because it has no real roots (its roots are i and -i). Quadratic polynomials with no real roots are irreducible in R[X].

Wrapping Up

So there you have it, guys! We successfully tackled the polynomial P(X), found its roots and their multiplicities, and decomposed it into irreducible polynomials in both C[X] and R[X]. We discovered that i is a root with multiplicity 2, 1 is a real root, and we expressed P(X) as (X - 1)(X - i)²(X + i)² in C[X] and as (X - 1)(X² + 1)² in R[X]. This process highlights the fundamental connection between the roots of a polynomial and its factorization. Understanding these concepts is crucial for more advanced topics in algebra and related fields. Keep practicing, and you'll become polynomial pros in no time!