Answer :
To determine which expression is a prime polynomial, we need to look at each option and see if it can be factored further over the integers.
A prime (or irreducible) polynomial is one that cannot be factored into simpler polynomials with integer coefficients.
Let's examine each option:
A. [tex]\(10x^4 - 5x^3 + 70x^2 + 3x\)[/tex]
First, factor out the greatest common factor, which is [tex]\(x\)[/tex]:
[tex]\[ x(10x^3 - 5x^2 + 70x + 3) \][/tex]
Since we could factor [tex]\(x\)[/tex] out, the polynomial is not prime.
B. [tex]\(x^3 - 27y^6\)[/tex]
This is a difference of cubes 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]:
[tex]\[ x^3 - (3y^2)^3 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4) \][/tex]
Since it factors, it's also not prime.
C. [tex]\(3x^2 + 18y\)[/tex]
Factor out the greatest common factor, which is 3:
[tex]\[ 3(x^2 + 6y) \][/tex]
Since we could factor 3 out, the polynomial is not prime.
D. [tex]\(x^4 + 20x^2 - 100\)[/tex]
This polynomial can be viewed as a quadratic in terms of [tex]\(x^2\)[/tex]:
Let [tex]\(u = x^2\)[/tex], then the expression becomes:
[tex]\[ u^2 + 20u - 100 \][/tex]
Trying to factor this quadratic:
We need two numbers that multiply to [tex]\(-100\)[/tex] and add to [tex]\(20\)[/tex], but no such integer pair exists. Thus, this polynomial cannot be factored over the integers.
Therefore, [tex]\(x^4 + 20x^2 - 100\)[/tex] is a prime polynomial.
Correct answer: D. [tex]\(x^4 + 20x^2 - 100\)[/tex]
A prime (or irreducible) polynomial is one that cannot be factored into simpler polynomials with integer coefficients.
Let's examine each option:
A. [tex]\(10x^4 - 5x^3 + 70x^2 + 3x\)[/tex]
First, factor out the greatest common factor, which is [tex]\(x\)[/tex]:
[tex]\[ x(10x^3 - 5x^2 + 70x + 3) \][/tex]
Since we could factor [tex]\(x\)[/tex] out, the polynomial is not prime.
B. [tex]\(x^3 - 27y^6\)[/tex]
This is a difference of cubes 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]:
[tex]\[ x^3 - (3y^2)^3 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4) \][/tex]
Since it factors, it's also not prime.
C. [tex]\(3x^2 + 18y\)[/tex]
Factor out the greatest common factor, which is 3:
[tex]\[ 3(x^2 + 6y) \][/tex]
Since we could factor 3 out, the polynomial is not prime.
D. [tex]\(x^4 + 20x^2 - 100\)[/tex]
This polynomial can be viewed as a quadratic in terms of [tex]\(x^2\)[/tex]:
Let [tex]\(u = x^2\)[/tex], then the expression becomes:
[tex]\[ u^2 + 20u - 100 \][/tex]
Trying to factor this quadratic:
We need two numbers that multiply to [tex]\(-100\)[/tex] and add to [tex]\(20\)[/tex], but no such integer pair exists. Thus, this polynomial cannot be factored over the integers.
Therefore, [tex]\(x^4 + 20x^2 - 100\)[/tex] is a prime polynomial.
Correct answer: D. [tex]\(x^4 + 20x^2 - 100\)[/tex]
