Question

Express x³ + 2x +3 as a product of irreducible polynomials in Z5[x].

Answer 1

To express x³ + 2x + 3 as a product of irreducible polynomials in Z5[x], we find that x = 2 and x = 4 are roots, and factoring them out we obtain (x - 2)(x - 4)Q(x). Polynomial division reveals that Q(x) is x² + 2x + 1, which is irreducible in Z5. Thus, the expression is (x - 2)(x - 4)(x² + 2x + 1) in Z5.

Step-by-step explanation:

To express the polynomial x³ + 2x + 3 as a product of irreducible polynomials in Z5[x], we must first find its roots in the field Z5, which consists of the elements {0, 1, 2, 3, 4}.

We will test each element of Z5 to see if it is a root of the polynomial by substituting x with each element and checking if the polynomial evaluates to 0 in Z5.

If x = 0, then f(0) = 0³ + 2×0 + 3 = 3 ≠ 0 in Z5.
If x = 1, then f(1) = 1³ + 2×1 + 3 = 1 + 2 + 3 = 6 ≡ 1 ≠ 0 in Z5.
If x = 2, then f(2) = 2³ + 2×2 + 3 = 8 + 4 + 3 = 15 ≡ 0 in Z5.
If x = 3, then f(3) = 3³ + 2×3 + 3 = 27 + 6 + 3 = 36 ≡ 1 ≠ 0 in Z5.
If x = 4, then f(4) = 4³ + 2×4 + 3 = 64 + 8 + 3 = 75 ≡ 0 in Z5.

Since 2 and 4 are roots of the polynomial, we can factor them out:

x³ + 2x + 3 = (x - 2)(x - 4)Q(x) in Z5

To find Q(x), we can perform polynomial division or use synthetic division:

(x³ + 0x² + 2x + 3) ÷ (x - 2) = x² + 2x + 1
Then we check if x² + 2x + 1 has roots in Z5:

If x = 0, 1, 3, or 4, it does not evaluate to 0; hence, x² + 2x + 1 is an irreducible polynomial in Z5.

Therefore, x³ + 2x + 3 can be expressed as (x - 2)(x - 4)(x² + 2x + 1) in Z5.

Answer 2

Yes, the polynomial
`$x^3+2 x+3$` is irreducible in
`\mathbb{Z}_5[x]`.

Check for Linear Factors (Roots):

- Directly evaluate the polynomial at each element of
`\mathbb{Z}_5` :

`\begin{aligned}& f(0)=3 \\eq 0 \\& f(1)=6 \equiv 1 \\eq 0 \\& f(2)=15 \equiv 0 \\eq 0 \\& f(3)=30 \equiv 0 \\eq 0 \\& f(4)=57 \equiv 2 \\eq 0\end{aligned}`

- Since the polynomial doesn't evaluate to 0 for any element in
`\mathbb{Z}_5`, it has no linear factors (roots) in
`\mathbb{Z}_5`.

Consider Factoring into Quadratic and Linear Factors:

- The only way the polynomial could be reducible is if it could be factored into a quadratic factor and a linear factor.

- However, in
`\mathbb{Z}_5[x]`, the only possible irreducible quadratics are of the form
`$x^2+a x+b$`, where
`a, b \in \mathbb{Z}_5` and
`$b \\eq 0$`.

- If
`$x^3+2 x+3$` were to factor as
`$\left(x^2+a x+b\right)(x+c)$`, then its constant term would be bc. But 3 cannot be written as the product of two nonzero elements in
`$\mathbb{Z}_5$`.

Therefore, the polynomial
`$x^3+2 x+3$` cannot be factored into any non-trivial factors in
`$\mathbb{Z}_5[x]$`, and thus it is irreducible.