Answer :
To determine which of the given polynomials is prime (irreducible), we need to analyze each polynomial to see if it can be factored into simpler polynomials with real coefficients. A prime polynomial is one that cannot be factored further over the set of real numbers.
Let's consider each polynomial one by one:
1. Polynomial: [tex]\( x^2 + x + 1 \)[/tex]
To check if this polynomial can be factored, we consider its discriminant [tex]\( \Delta \)[/tex] given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = 1 \)[/tex].
Substituting these values in, we get:
[tex]\[ \Delta = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \][/tex]
Since the discriminant is negative ([tex]\(\Delta < 0\)[/tex]), the polynomial has no real roots and cannot be factored over the reals. Therefore, [tex]\( x^2 + x + 1 \)[/tex] is irreducible (prime).
2. Polynomial: [tex]\( x^2 - x - 2 \)[/tex]
Again, we check the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -2 \)[/tex].
Substituting these in, we get:
[tex]\[ \Delta = (-1)^2 - 4 \cdot 1 \cdot (-2) = 1 + 8 = 9 \][/tex]
Since the discriminant is positive ([tex]\(\Delta > 0\)[/tex]), the polynomial has real roots and can be factored. Specifically, it can be factored as:
[tex]\[ x^2 - x - 2 = (x - 2)(x + 1) \][/tex]
Therefore, [tex]\( x^2 - x - 2 \)[/tex] is not prime.
3. Polynomial: [tex]\( x^2 - 12x + 11 \)[/tex]
We check the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -12 \)[/tex], and [tex]\( c = 11 \)[/tex].
Substituting these in, we get:
[tex]\[ \Delta = (-12)^2 - 4 \cdot 1 \cdot 11 = 144 - 44 = 100 \][/tex]
Since the discriminant is positive ([tex]\(\Delta > 0\)[/tex]), the polynomial has real roots and can be factored. Specifically, it can be factored as:
[tex]\[ x^2 - 12x + 11 = (x - 1)(x - 11) \][/tex]
Therefore, [tex]\( x^2 - 12x + 11 \)[/tex] is not prime.
4. Polynomial: [tex]\( x^2 + 2x - 8 \)[/tex]
We check the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -8 \)[/tex].
Substituting these in, we get:
[tex]\[ \Delta = 2^2 - 4 \cdot 1 \cdot (-8) = 4 + 32 = 36 \][/tex]
Since the discriminant is positive ([tex]\(\Delta > 0\)[/tex]), the polynomial has real roots and can be factored. Specifically, it can be factored as:
[tex]\[ x^2 + 2x - 8 = (x + 4)(x - 2) \][/tex]
Therefore, [tex]\( x^2 + 2x - 8 \)[/tex] is not prime.
In conclusion, the polynomial that is prime (irreducible) among the given choices is:
[tex]\[ \boxed{1 \text{ (i.e., } x^2 + x + 1\text{)}} \][/tex]
Let's consider each polynomial one by one:
1. Polynomial: [tex]\( x^2 + x + 1 \)[/tex]
To check if this polynomial can be factored, we consider its discriminant [tex]\( \Delta \)[/tex] given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = 1 \)[/tex].
Substituting these values in, we get:
[tex]\[ \Delta = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \][/tex]
Since the discriminant is negative ([tex]\(\Delta < 0\)[/tex]), the polynomial has no real roots and cannot be factored over the reals. Therefore, [tex]\( x^2 + x + 1 \)[/tex] is irreducible (prime).
2. Polynomial: [tex]\( x^2 - x - 2 \)[/tex]
Again, we check the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -2 \)[/tex].
Substituting these in, we get:
[tex]\[ \Delta = (-1)^2 - 4 \cdot 1 \cdot (-2) = 1 + 8 = 9 \][/tex]
Since the discriminant is positive ([tex]\(\Delta > 0\)[/tex]), the polynomial has real roots and can be factored. Specifically, it can be factored as:
[tex]\[ x^2 - x - 2 = (x - 2)(x + 1) \][/tex]
Therefore, [tex]\( x^2 - x - 2 \)[/tex] is not prime.
3. Polynomial: [tex]\( x^2 - 12x + 11 \)[/tex]
We check the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = -12 \)[/tex], and [tex]\( c = 11 \)[/tex].
Substituting these in, we get:
[tex]\[ \Delta = (-12)^2 - 4 \cdot 1 \cdot 11 = 144 - 44 = 100 \][/tex]
Since the discriminant is positive ([tex]\(\Delta > 0\)[/tex]), the polynomial has real roots and can be factored. Specifically, it can be factored as:
[tex]\[ x^2 - 12x + 11 = (x - 1)(x - 11) \][/tex]
Therefore, [tex]\( x^2 - 12x + 11 \)[/tex] is not prime.
4. Polynomial: [tex]\( x^2 + 2x - 8 \)[/tex]
We check the discriminant:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
where [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -8 \)[/tex].
Substituting these in, we get:
[tex]\[ \Delta = 2^2 - 4 \cdot 1 \cdot (-8) = 4 + 32 = 36 \][/tex]
Since the discriminant is positive ([tex]\(\Delta > 0\)[/tex]), the polynomial has real roots and can be factored. Specifically, it can be factored as:
[tex]\[ x^2 + 2x - 8 = (x + 4)(x - 2) \][/tex]
Therefore, [tex]\( x^2 + 2x - 8 \)[/tex] is not prime.
In conclusion, the polynomial that is prime (irreducible) among the given choices is:
[tex]\[ \boxed{1 \text{ (i.e., } x^2 + x + 1\text{)}} \][/tex]
