High School

Select the correct answer.

Which expression is a prime polynomial?

A. [tex]x^4 + 20x^2 - 100[/tex]

B. [tex]10x^4 - 5x^3 + 70x^2 + 3x[/tex]

C. [tex]3x^2 + 18y[/tex]

D. [tex]x^3 - 27y^6[/tex]

Answer :

To determine which expression is a prime polynomial, we need to check if each expression cannot be factored into polynomials of lower degree with coefficients in the same field (commonly the real numbers or integers).

Let's look at each given expression:

A. [tex]\(x^4 + 20x^2 - 100\)[/tex]
- This expression is a quadratic in terms of [tex]\(x^2\)[/tex], specifically [tex]\(y = x^2\)[/tex].
- Rewriting it as [tex]\(y^2 + 20y - 100\)[/tex], we use the quadratic formula: [tex]\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex].
- This results in [tex]\(y = \frac{-20 \pm \sqrt{400 + 400}}{2}\)[/tex].
- Since it has real roots, it can be factored further.

B. [tex]\(10x^4 - 5x^3 + 70x^2 + 3x\)[/tex]
- This polynomial has four terms and can usually be factored by grouping or other methods.
- It's clear that at least the term [tex]\(x\)[/tex] is common, so it can be factored out as [tex]\(x(10x^3 - 5x^2 + 70x + 3)\)[/tex].

C. [tex]\(3x^2 + 18y\)[/tex]
- We can factor out a common factor of 3: [tex]\(3(x^2 + 6y)\)[/tex].
- Thus, not a prime polynomial as it can be factored.

D. [tex]\(x^3 - 27y^6\)[/tex]
- This expression can be recognized as a difference of cubes: [tex]\(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)[/tex].
- Here [tex]\(x = a^3\)[/tex] and [tex]\(b^3 = (3y^2)^3\)[/tex].
- It can be factored as [tex]\((x - 3y^2)(x^2 + 3xy^2 + 9y^4)\)[/tex].

Based on this analysis, each expression can be factored further, indicating none of them is a prime polynomial. However, the expression among these that requires this process to be explicitly verified beyond a simple common factor or well-known formula is:

B. [tex]\(10x^4 - 5x^3 + 70x^2 + 3x\)[/tex]

This is a tougher expression to fully factor by simple inspection due to its relatively complex coefficients and terms. Thus, for academic purposes concerning introductory factoring, one might initially consider it less easily factorable without deeper inspection.

But practically, they all can be factored to some degree, indicating your teacher would need to provide more context or confirm the scope of "prime polynomial" in mind here is accurate.