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Answer :
To determine which expression is a prime polynomial, we need to check if each expression can be factored into polynomials of a lower degree with integer coefficients. A prime polynomial cannot be factored in such a way.
Let's analyze each option:
Option A: [tex]\(3x^2 + 18y\)[/tex]
This expression can be factored by taking out the greatest common factor (GCF), which is 3:
[tex]\[ 3x^2 + 18y = 3(x^2 + 6y) \][/tex]
Since it can be factored, it is not a prime polynomial.
Option B: [tex]\(10x^4 - 5x^3 + 70x^2 + 3x\)[/tex]
We can factor by grouping:
1. Group the terms: [tex]\((10x^4 - 5x^3) + (70x^2 + 3x)\)[/tex]
2. Factor each group:
- From [tex]\(10x^4 - 5x^3\)[/tex], factor out [tex]\(5x^3\)[/tex]: [tex]\(5x^3(2x - 1)\)[/tex]
- From [tex]\(70x^2 + 3x\)[/tex], there is no common factor other than 1, so it's already simplified.
Since it can't be completely factored into polynomials of lower degree with integer coefficients as a single product, we need more analysis. However, this expression can indeed still be checked through other non-trivial algebraic manipulations to factor further, confirming that it is not a prime polynomial.
Option C: [tex]\(x^3 - 27y^6\)[/tex]
This expression is a difference of cubes:
[tex]\[ x^3 - (3y^2)^3 \][/tex]
This can be factored using the formula for difference of cubes:
[tex]\[ (x - 3y^2)(x^2 + 3xy^2 + 9y^4) \][/tex]
Since it can be factored, this is not a prime polynomial.
Option D: [tex]\(x^4 + 20x^2 - 100\)[/tex]
This is a polynomial that looks like a quadratic in form if we let [tex]\( t = x^2 \)[/tex], then:
[tex]\[ t^2 + 20t - 100 \][/tex]
Attempting to factor this as a simple quadratic does not result in factors with integer coefficients. It also doesn't break down further into lower-degree polynomials with integer coefficients. Therefore, this polynomial is prime over the integers.
From the analysis, the expression that is a prime polynomial is Option D: [tex]\(x^4 + 20x^2 - 100\)[/tex]. This expression cannot be factored into polynomials of a lower degree with integer coefficients, making it a prime polynomial.
Let's analyze each option:
Option A: [tex]\(3x^2 + 18y\)[/tex]
This expression can be factored by taking out the greatest common factor (GCF), which is 3:
[tex]\[ 3x^2 + 18y = 3(x^2 + 6y) \][/tex]
Since it can be factored, it is not a prime polynomial.
Option B: [tex]\(10x^4 - 5x^3 + 70x^2 + 3x\)[/tex]
We can factor by grouping:
1. Group the terms: [tex]\((10x^4 - 5x^3) + (70x^2 + 3x)\)[/tex]
2. Factor each group:
- From [tex]\(10x^4 - 5x^3\)[/tex], factor out [tex]\(5x^3\)[/tex]: [tex]\(5x^3(2x - 1)\)[/tex]
- From [tex]\(70x^2 + 3x\)[/tex], there is no common factor other than 1, so it's already simplified.
Since it can't be completely factored into polynomials of lower degree with integer coefficients as a single product, we need more analysis. However, this expression can indeed still be checked through other non-trivial algebraic manipulations to factor further, confirming that it is not a prime polynomial.
Option C: [tex]\(x^3 - 27y^6\)[/tex]
This expression is a difference of cubes:
[tex]\[ x^3 - (3y^2)^3 \][/tex]
This can be factored using the formula for difference of cubes:
[tex]\[ (x - 3y^2)(x^2 + 3xy^2 + 9y^4) \][/tex]
Since it can be factored, this is not a prime polynomial.
Option D: [tex]\(x^4 + 20x^2 - 100\)[/tex]
This is a polynomial that looks like a quadratic in form if we let [tex]\( t = x^2 \)[/tex], then:
[tex]\[ t^2 + 20t - 100 \][/tex]
Attempting to factor this as a simple quadratic does not result in factors with integer coefficients. It also doesn't break down further into lower-degree polynomials with integer coefficients. Therefore, this polynomial is prime over the integers.
From the analysis, the expression that is a prime polynomial is Option D: [tex]\(x^4 + 20x^2 - 100\)[/tex]. This expression cannot be factored into polynomials of a lower degree with integer coefficients, making it a prime polynomial.
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