Answer :
Sure! Let's determine which of the given expressions is a prime polynomial. A prime polynomial is one that cannot be factored into polynomials of lower degree with integer coefficients.
Let's analyze each option:
Option A: [tex]\(3x^2 + 18y\)[/tex]
- This expression can be factored by taking the greatest common factor (GCF), which is 3.
- Factoring gives: [tex]\(3(x^2 + 6y)\)[/tex].
- Since it can be factored, it is not a prime polynomial.
Option B: [tex]\(x^3 - 27y^6\)[/tex]
- This is a difference of cubes, which follows the formula: [tex]\(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)[/tex].
- Here, [tex]\(a = x\)[/tex] and [tex]\(b = 3y^2\)[/tex].
- Factoring gives: [tex]\((x - 3y^2)(x^2 + 3xy^2 + 9y^4)\)[/tex].
- Since it can be factored, it is not a prime polynomial.
Option C: [tex]\(10x^4 - 5x^3 + 70x^2 + 3x\)[/tex]
- We can factor out the greatest common factor, which is [tex]\(x\)[/tex].
- Factoring gives: [tex]\(x(10x^3 - 5x^2 + 70x + 3)\)[/tex].
- Since it can be factored, it is not a prime polynomial.
Option D: [tex]\(x^4 + 20x^2 - 100\)[/tex]
- Let's try to factor this expression. Notice that it resembles a quadratic form where we can let [tex]\(z = x^2\)[/tex], reducing it to: [tex]\(z^2 + 20z - 100\)[/tex].
- We attempt to factor this like a quadratic:
- The factors of 100 are 10 and -10, which add up to zero, not 20.
- Or other possibilities with integer factors won't give a sum of 20 directly.
- The test for factorization doesn't yield integer factors, and breaking it further with complex factorization shows it can't be reduced cleanly with integer components.
- Therefore, it is very likely prime over the integers.
The expression that is a prime polynomial is Option D: [tex]\(x^4 + 20x^2 - 100\)[/tex].
Let's analyze each option:
Option A: [tex]\(3x^2 + 18y\)[/tex]
- This expression can be factored by taking the greatest common factor (GCF), which is 3.
- Factoring gives: [tex]\(3(x^2 + 6y)\)[/tex].
- Since it can be factored, it is not a prime polynomial.
Option B: [tex]\(x^3 - 27y^6\)[/tex]
- This is a difference of cubes, which follows the formula: [tex]\(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)[/tex].
- Here, [tex]\(a = x\)[/tex] and [tex]\(b = 3y^2\)[/tex].
- Factoring gives: [tex]\((x - 3y^2)(x^2 + 3xy^2 + 9y^4)\)[/tex].
- Since it can be factored, it is not a prime polynomial.
Option C: [tex]\(10x^4 - 5x^3 + 70x^2 + 3x\)[/tex]
- We can factor out the greatest common factor, which is [tex]\(x\)[/tex].
- Factoring gives: [tex]\(x(10x^3 - 5x^2 + 70x + 3)\)[/tex].
- Since it can be factored, it is not a prime polynomial.
Option D: [tex]\(x^4 + 20x^2 - 100\)[/tex]
- Let's try to factor this expression. Notice that it resembles a quadratic form where we can let [tex]\(z = x^2\)[/tex], reducing it to: [tex]\(z^2 + 20z - 100\)[/tex].
- We attempt to factor this like a quadratic:
- The factors of 100 are 10 and -10, which add up to zero, not 20.
- Or other possibilities with integer factors won't give a sum of 20 directly.
- The test for factorization doesn't yield integer factors, and breaking it further with complex factorization shows it can't be reduced cleanly with integer components.
- Therefore, it is very likely prime over the integers.
The expression that is a prime polynomial is Option D: [tex]\(x^4 + 20x^2 - 100\)[/tex].
