Answer :
To determine which expression is a prime polynomial, we need to understand what a prime polynomial is. A prime polynomial is one that cannot be factored into the product of two non-constant polynomials with integer coefficients.
Let's examine each option:
A. [tex]\(3x^2 + 18y\)[/tex]
- This polynomial can be factored. We can take out a common factor of 3:
[tex]\[
3x^2 + 18y = 3(x^2 + 6y)
\][/tex]
- Since it can be factored further, it is not a prime polynomial.
B. [tex]\(10x^4 - 5x^3 + 70x^2 + 3x\)[/tex]
- This polynomial seems complicated and is reducible. It can be factored, so it is not a prime polynomial.
C. [tex]\(x^3 - 27y^6\)[/tex]
- This is a difference of cubes:
[tex]\[
x^3 - (3y^2)^3 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4)
\][/tex]
- It can be factored using the difference of cubes formula, so it is not a prime polynomial.
D. [tex]\(x^4 + 20x^2 - 100\)[/tex]
- This polynomial can be viewed as a quadratic in form if we substitute [tex]\(z = x^2\)[/tex]:
[tex]\[
z^2 + 20z - 100
\][/tex]
- This quadratic can be factored further, showing it is not a prime polynomial.
Based on the examination of these options:
- Option A: It can be factored and is not a prime polynomial.
- Option B: It can be factored and is not a prime polynomial.
- Option C: It can be factored as a difference of cubes and is not a prime polynomial.
- Option D: It can be factored as a quadratic and is not a prime polynomial.
From these analyses, none of the polynomials in the options are prime. However, if you were given that option B returns true that it is a prime polynomial and others are not, then based on this logic, option A is indeed prime.
Let's examine each option:
A. [tex]\(3x^2 + 18y\)[/tex]
- This polynomial can be factored. We can take out a common factor of 3:
[tex]\[
3x^2 + 18y = 3(x^2 + 6y)
\][/tex]
- Since it can be factored further, it is not a prime polynomial.
B. [tex]\(10x^4 - 5x^3 + 70x^2 + 3x\)[/tex]
- This polynomial seems complicated and is reducible. It can be factored, so it is not a prime polynomial.
C. [tex]\(x^3 - 27y^6\)[/tex]
- This is a difference of cubes:
[tex]\[
x^3 - (3y^2)^3 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4)
\][/tex]
- It can be factored using the difference of cubes formula, so it is not a prime polynomial.
D. [tex]\(x^4 + 20x^2 - 100\)[/tex]
- This polynomial can be viewed as a quadratic in form if we substitute [tex]\(z = x^2\)[/tex]:
[tex]\[
z^2 + 20z - 100
\][/tex]
- This quadratic can be factored further, showing it is not a prime polynomial.
Based on the examination of these options:
- Option A: It can be factored and is not a prime polynomial.
- Option B: It can be factored and is not a prime polynomial.
- Option C: It can be factored as a difference of cubes and is not a prime polynomial.
- Option D: It can be factored as a quadratic and is not a prime polynomial.
From these analyses, none of the polynomials in the options are prime. However, if you were given that option B returns true that it is a prime polynomial and others are not, then based on this logic, option A is indeed prime.
