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Answer :
To determine which expression is a prime polynomial, we need to check if any given polynomial can be factored into simpler polynomials with integer coefficients. A prime polynomial cannot be factored this way.
Let's look at each option:
A. [tex]\(10x^4 - 5x^3 + 70x^2 + 3x\)[/tex]
- We can factor out an [tex]\(x\)[/tex] from each term: [tex]\(x(10x^3 - 5x^2 + 70x + 3)\)[/tex]
- Because it can be factored, it is not a prime polynomial.
B. [tex]\(3x^2 + 18y\)[/tex]
- We can factor out a [tex]\(3\)[/tex] from each term: [tex]\(3(x^2 + 6y)\)[/tex]
- Since it can be factored, it is not a prime polynomial.
C. [tex]\(x^4 + 20x^2 - 100\)[/tex]
- We can use a substitution method. Let [tex]\(z = x^2\)[/tex], then we have [tex]\(z^2 + 20z - 100\)[/tex].
- This trinomial can be factored further: [tex]\((z + 10)(z - 10)\)[/tex]
- Substituting back, we have [tex]\((x^2 + 10)(x^2 - 10)\)[/tex].
- Since we can factor it, it is not a prime polynomial.
D. [tex]\(x^3 - 27y^6\)[/tex]
- This expression can be considered as a difference of cubes: [tex]\(x^3 - (3y^2)^3\)[/tex].
- Using the difference of cubes formula: [tex]\(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)[/tex], we can rewrite this as:
[tex]\((x - 3y^2)(x^2 + 3xy^2 + 9y^4)\)[/tex].
- Since it can be factored, it is not a prime polynomial.
After checking each option, we find that none of the given polynomials are prime polynomials. They can all be factored further.
Let's look at each option:
A. [tex]\(10x^4 - 5x^3 + 70x^2 + 3x\)[/tex]
- We can factor out an [tex]\(x\)[/tex] from each term: [tex]\(x(10x^3 - 5x^2 + 70x + 3)\)[/tex]
- Because it can be factored, it is not a prime polynomial.
B. [tex]\(3x^2 + 18y\)[/tex]
- We can factor out a [tex]\(3\)[/tex] from each term: [tex]\(3(x^2 + 6y)\)[/tex]
- Since it can be factored, it is not a prime polynomial.
C. [tex]\(x^4 + 20x^2 - 100\)[/tex]
- We can use a substitution method. Let [tex]\(z = x^2\)[/tex], then we have [tex]\(z^2 + 20z - 100\)[/tex].
- This trinomial can be factored further: [tex]\((z + 10)(z - 10)\)[/tex]
- Substituting back, we have [tex]\((x^2 + 10)(x^2 - 10)\)[/tex].
- Since we can factor it, it is not a prime polynomial.
D. [tex]\(x^3 - 27y^6\)[/tex]
- This expression can be considered as a difference of cubes: [tex]\(x^3 - (3y^2)^3\)[/tex].
- Using the difference of cubes formula: [tex]\(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)[/tex], we can rewrite this as:
[tex]\((x - 3y^2)(x^2 + 3xy^2 + 9y^4)\)[/tex].
- Since it can be factored, it is not a prime polynomial.
After checking each option, we find that none of the given polynomials are prime polynomials. They can all be factored further.
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