Answer :
Let's analyze each polynomial to determine if it is a prime polynomial, which means it cannot be factored into the product of two non-constant polynomials with coefficients in the same field.
### Option A: [tex]\( 10x^4 - 5x^3 + 70x^2 + 3x \)[/tex]
First, let's look for the greatest common factor (GCF) of all the terms:
- The GCF is [tex]\( x \)[/tex].
Factor out the GCF:
[tex]\[ 10x^4 - 5x^3 + 70x^2 + 3x = x(10x^3 - 5x^2 + 70x + 3) \][/tex]
Since it can be factored, [tex]\( 10x^4 - 5x^3 + 70x^2 + 3x \)[/tex] is not a prime polynomial.
### Option B: [tex]\( x^4 + 20x^2 - 100 \)[/tex]
Check if it can be factored as a product of simpler polynomials:
- Substitute [tex]\( y = x^2 \)[/tex] to simplify:
[tex]\[ y^2 + 20y - 100 \][/tex]
Look for factors of [tex]\(-100\)[/tex] that add up to 20:
- The quadratic cannot be factored easily into simpler integer factors.
Put back [tex]\( y = x^2 \)[/tex] and recheck factorization:
- Since we cannot factor the quadratic form [tex]\( y^2 + 20y - 100 \)[/tex], it remains in this form if [tex]\( x^4 + 20x^2 - 100 \)[/tex] is irreducible.
### Option C: [tex]\( 3x^2 + 18y \)[/tex]
Look for the greatest common factor (GCF) of [tex]\( 3x^2 \)[/tex] and [tex]\( 18y \)[/tex]:
- The GCF is [tex]\( 3 \)[/tex].
Factor out the GCF:
[tex]\[ 3x^2 + 18y = 3(x^2 + 6y) \][/tex]
Since it can be factored, [tex]\( 3x^2 + 18y \)[/tex] is not a prime polynomial.
### Option D: [tex]\( x^3 - 27y^6 \)[/tex]
This expression looks like a difference of cubes:
[tex]\[ x^3 - 27y^6 = x^3 - (3y^2)^3 \][/tex]
Using the difference of cubes formula, [tex]\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)[/tex], where [tex]\( a = x \)[/tex] and [tex]\( b = 3y^2 \)[/tex]:
[tex]\[ x^3 - 27y^6 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4) \][/tex]
Since it can be factored, [tex]\( x^3 - 27y^6 \)[/tex] is not a prime polynomial.
### Conclusion
Among the given options, only [tex]\( x^4 + 20x^2 - 100 \)[/tex] cannot be factored further using elementary algebraic techniques, indicating it is a prime polynomial.
Thus, the correct answer is:
[tex]\[ \boxed{B} \][/tex]
### Option A: [tex]\( 10x^4 - 5x^3 + 70x^2 + 3x \)[/tex]
First, let's look for the greatest common factor (GCF) of all the terms:
- The GCF is [tex]\( x \)[/tex].
Factor out the GCF:
[tex]\[ 10x^4 - 5x^3 + 70x^2 + 3x = x(10x^3 - 5x^2 + 70x + 3) \][/tex]
Since it can be factored, [tex]\( 10x^4 - 5x^3 + 70x^2 + 3x \)[/tex] is not a prime polynomial.
### Option B: [tex]\( x^4 + 20x^2 - 100 \)[/tex]
Check if it can be factored as a product of simpler polynomials:
- Substitute [tex]\( y = x^2 \)[/tex] to simplify:
[tex]\[ y^2 + 20y - 100 \][/tex]
Look for factors of [tex]\(-100\)[/tex] that add up to 20:
- The quadratic cannot be factored easily into simpler integer factors.
Put back [tex]\( y = x^2 \)[/tex] and recheck factorization:
- Since we cannot factor the quadratic form [tex]\( y^2 + 20y - 100 \)[/tex], it remains in this form if [tex]\( x^4 + 20x^2 - 100 \)[/tex] is irreducible.
### Option C: [tex]\( 3x^2 + 18y \)[/tex]
Look for the greatest common factor (GCF) of [tex]\( 3x^2 \)[/tex] and [tex]\( 18y \)[/tex]:
- The GCF is [tex]\( 3 \)[/tex].
Factor out the GCF:
[tex]\[ 3x^2 + 18y = 3(x^2 + 6y) \][/tex]
Since it can be factored, [tex]\( 3x^2 + 18y \)[/tex] is not a prime polynomial.
### Option D: [tex]\( x^3 - 27y^6 \)[/tex]
This expression looks like a difference of cubes:
[tex]\[ x^3 - 27y^6 = x^3 - (3y^2)^3 \][/tex]
Using the difference of cubes formula, [tex]\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)[/tex], where [tex]\( a = x \)[/tex] and [tex]\( b = 3y^2 \)[/tex]:
[tex]\[ x^3 - 27y^6 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4) \][/tex]
Since it can be factored, [tex]\( x^3 - 27y^6 \)[/tex] is not a prime polynomial.
### Conclusion
Among the given options, only [tex]\( x^4 + 20x^2 - 100 \)[/tex] cannot be factored further using elementary algebraic techniques, indicating it is a prime polynomial.
Thus, the correct answer is:
[tex]\[ \boxed{B} \][/tex]
