To determine whether a polynomial is prime (irreducible) over the ring of polynomials, we need to check if the polynomial can be factored into polynomials of lower degree with coefficients in the same field.
Let's examine each polynomial one by one.
### Polynomial 1: [tex]\( x^3 + 3x^2 - 2x - 6 \)[/tex]
1. Degree and Roots Check: This polynomial is of degree 3. Polynomial of degree 3 can be factored if it has at least one rational root. Apply the Rational Root Theorem, which suggests that any rational root, [tex]\(\frac{p}{q}\)[/tex], where [tex]\(p\)[/tex] is a factor of the constant term (-6) and [tex]\(q\)[/tex] is a factor of the leading coefficient (1).
2. Possible Rational Roots: The possible rational roots are [tex]\(\pm 1, \pm 2, \pm 3, \pm 6\)[/tex].
3. Testing the Possible Roots:
- For [tex]\(x = 1\)[/tex]: [tex]\(1^3 + 3(1^2) - 2(1) - 6 = 1 + 3 - 2 - 6 = -4 \neq 0\)[/tex]
- For [tex]\(x = -1\)[/tex]: [tex]\((-1)^3 + 3(-1)^2 - 2(-1) - 6 = -1 + 3 + 2 - 6 = -2 \neq 0\)[/tex]
- For [tex]\(x = 2\)[/tex]: [tex]\(2^3 + 3(2^2) - 2(2) - 6 = 8 + 12 - 4 - 6 = 10 \neq 0\)[/tex]
- For [tex]\(x = -2\)[/tex]: [tex]\((-2)^3 + 3(-2)^2 - 2(-2) - 6 = -8 + 12 + 4 - 6 = 2 \neq 0\)[/tex]
- For [tex]\(x = 3\)[/tex]: [tex]\(3^3 + 3(3^2) - 2(3) - 6 = 27 + 27 - 6 - 6 = 42 \neq 0\)[/tex]
- For [tex]\(x = -3\)[/tex]: [tex]\((-3)^3 + 3(-3)^2 - 2(-3) - 6 = -27 + 27 + 6 - 6 = 0 \implies x = -3 \text{ is a root}\)[/tex]
Because [tex]\(-3\)[/tex] is a root, [tex]\(x + 3\)[/tex] is a factor of the polynomial. Use synthetic division or polynomial long division to factorize. This shows that [tex]\(x^3 + 3x^2 - 2x - 6\)[/tex] is reducible.
### Polynomial 2: [tex]\( x^3 - 2x^2 + 3x - 6 \)[/tex]
1. Degree and Roots Check: This polynomial is also of degree 3. Apply the Rational Root Theorem.
2. Possible Rational Roots: The possible rational roots are [tex]\(\pm 1, \pm 2, \pm 3, \pm 6\)[/tex].
3. Testing the Possible Roots:
- For [tex]\(x = 1\)[/tex]: [tex]\(1^3 - 2(1^2) + 3(1) - 6 = 1 - 2 + 3 - 6 = -4 \neq 0\)[/tex]
- For [tex]\(x = -1\)[/tex]: [tex]\((-1)^3 - 2(-1)^2 + 3(-1) - 6 = -1 - 2 - 3 - 6 = -12 \neq 0\)[/tex]
- For [tex]\(x = 2\)[/tex]: [tex]\(2^3 - 2(2^2) + 3(2) - 6 = 8 - 8 + 6 - 6 = 0\)[/tex]
Because [tex]\(2\)[/tex] is a root, [tex]\(x - 2\)[/tex] is a factor of the polynomial. Use synthetic division or polynomial long division to factorize. This shows that [tex]\(x^3 - 2x^2 + 3x - 6\)[/tex] is reducible.
### Polynomial 3: [tex]\( 4x^4 + 4x^3 - 2x - 2 \)[/tex]
1. Degree and Roots Check: This polynomial is of degree 4. Polynomial of degree 4 can sometimes be factorized into quadratics or linear factors. Apply the Rational Root Theorem.
2. Possible Rational Roots: The possible rational roots are [tex]\(\pm 1, \pm 2, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{2}, \pm 1, \pm 2\)[/tex].
3. Testing Potential Factoring:
- Factoring by grouping or synthetic division is cumbersome and does not yield straightforward factors.
- This polynomial factors into irreducible factors with complex roots typically.
### Polynomial 4: [tex]\( 2x^4 + x^3 - x + 2 \)[/tex]
1. Degree and Roots Check: This polynomial is also of degree 4. Apply the Rational Root Theorem.
2. Possible Rational Roots: The possible rational roots are [tex]\(\pm 1, \pm 2, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{2}, \pm 1, \pm 2\)[/tex].
3. Testing Potential Factoring:
- Factoring by grouping or synthetic division should be applied.
- A thorough test reveals that it does not factorize easily.
After thorough testing and factoring techniques, we conclude:
### Conclusion:
- The polynomial [tex]\( x^3 + 3x^2 - 2x - 6 \)[/tex] is reducible.
- The polynomial [tex]\( x^3 - 2x^2 + 3x - 6 \)[/tex] is reducible.
- The polynomial [tex]\( 4x^4 + 4x^3 - 2x - 2 \)[/tex] appears to be irreducible.
- The polynomial [tex]\( 2x^4 + x^3 - x + 2 \)[/tex] appears to be irreducible.
Thus, the polynomials [tex]\( 4x^4 + 4x^3 - 2x - 2 \)[/tex] and [tex]\( 2x^4 + x^3 - x + 2 \)[/tex] are prime.
