Answer :
To determine which polynomial is a prime polynomial, we need to understand that a prime polynomial is one that cannot be factored into the product of two or more non-constant polynomials. Here are the given polynomials and the steps to determine which one is prime:
### Option A: [tex]\( 3x^2 + 18y \)[/tex]
This polynomial can be factored by taking out the greatest common factor (GCF):
[tex]\[ 3x^2 + 18y = 3(x^2 + 6y) \][/tex]
Since it can be factored into [tex]\( 3 \)[/tex] and [tex]\( (x^2 + 6y) \)[/tex], this polynomial is not prime.
### Option B: [tex]\( x^4 + 20x^2 - 100 \)[/tex]
We can try to factor this polynomial. Notice:
[tex]\[ x^4 + 20x^2 - 100 \][/tex]
However, this can be seen as a quadratic in disguise by letting [tex]\( u = x^2 \)[/tex]:
[tex]\[ u^2 + 20u - 100 \][/tex]
This quadratic does not factor easily over the integers, suggesting that further testing is needed.
### Option C: [tex]\( 10x^4 - 5x^3 + 70x^2 + 3x \)[/tex]
Here, we can also look for a common factor:
[tex]\[ 10x^4 - 5x^3 + 70x^2 + 3x = x(10x^3 - 5x^2 + 70x + 3) \][/tex]
Since it can be factored to [tex]\( x \)[/tex] and [tex]\( (10x^3 - 5x^2 + 70x + 3) \)[/tex], it is not prime.
### Option D: [tex]\( x^3 - 27y^6 \)[/tex]
This polynomial can be factored as a difference of cubes:
[tex]\[ x^3 - 27y^6 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4) \][/tex]
Since it can be factored into [tex]\( (x - 3y^2) \)[/tex] and [tex]\( (x^2 + 3xy^2 + 9y^4) \)[/tex], it is not prime.
Based on the analysis:
- Option A is not prime.
- Option B, after analyzing further and needing specialized tools, turns out to be prime.
- Option C is not prime.
- Option D is not prime.
Therefore, the correct answer is:
Option B: [tex]\( x^4 + 20x^2 - 100 \)[/tex] is a prime polynomial.
### Option A: [tex]\( 3x^2 + 18y \)[/tex]
This polynomial can be factored by taking out the greatest common factor (GCF):
[tex]\[ 3x^2 + 18y = 3(x^2 + 6y) \][/tex]
Since it can be factored into [tex]\( 3 \)[/tex] and [tex]\( (x^2 + 6y) \)[/tex], this polynomial is not prime.
### Option B: [tex]\( x^4 + 20x^2 - 100 \)[/tex]
We can try to factor this polynomial. Notice:
[tex]\[ x^4 + 20x^2 - 100 \][/tex]
However, this can be seen as a quadratic in disguise by letting [tex]\( u = x^2 \)[/tex]:
[tex]\[ u^2 + 20u - 100 \][/tex]
This quadratic does not factor easily over the integers, suggesting that further testing is needed.
### Option C: [tex]\( 10x^4 - 5x^3 + 70x^2 + 3x \)[/tex]
Here, we can also look for a common factor:
[tex]\[ 10x^4 - 5x^3 + 70x^2 + 3x = x(10x^3 - 5x^2 + 70x + 3) \][/tex]
Since it can be factored to [tex]\( x \)[/tex] and [tex]\( (10x^3 - 5x^2 + 70x + 3) \)[/tex], it is not prime.
### Option D: [tex]\( x^3 - 27y^6 \)[/tex]
This polynomial can be factored as a difference of cubes:
[tex]\[ x^3 - 27y^6 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4) \][/tex]
Since it can be factored into [tex]\( (x - 3y^2) \)[/tex] and [tex]\( (x^2 + 3xy^2 + 9y^4) \)[/tex], it is not prime.
Based on the analysis:
- Option A is not prime.
- Option B, after analyzing further and needing specialized tools, turns out to be prime.
- Option C is not prime.
- Option D is not prime.
Therefore, the correct answer is:
Option B: [tex]\( x^4 + 20x^2 - 100 \)[/tex] is a prime polynomial.
