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Answer :
Sure! Let's look into the given expressions one by one to determine which one is a prime polynomial. A prime polynomial is one that cannot be factored further over the set of integers.
A. [tex]\(x^3 - 27y^6\)[/tex]
This expression can be factored because it is a difference of cubes. The formula for the difference of cubes is [tex]\(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)[/tex]. Here, [tex]\(x^3\)[/tex] is [tex]\(a^3\)[/tex] and [tex]\( (3y^2)^3\)[/tex] is [tex]\(b^3\)[/tex]. So, this can be factored and is not a prime polynomial.
B. [tex]\(3x^2 + 18y\)[/tex]
This expression can be factored by taking out the greatest common factor (GCF). The GCF in this case is 3, so it can be rewritten as [tex]\(3(x^2 + 6y)\)[/tex]. Since it can be factored, it is not a prime polynomial.
C. [tex]\(10x^4 - 5x^3 + 70x^2 + 3x\)[/tex]
We can factor out an [tex]\(x\)[/tex] as a common factor, resulting in [tex]\(x(10x^3 - 5x^2 + 70x + 3)\)[/tex]. Since it can be factored further, it is not a prime polynomial.
D. [tex]\(x^4 + 20x^2 - 100\)[/tex]
This polynomial is a bit more challenging. We can try to factor this as a quadratic in form. Let [tex]\(z = x^2\)[/tex]. Then, the expression becomes [tex]\(z^2 + 20z - 100\)[/tex], which is quadratic in [tex]\(z\)[/tex].
Let's use the quadratic formula [tex]\(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] to check the roots:
- Here, [tex]\(a = 1\)[/tex], [tex]\(b = 20\)[/tex], and [tex]\(c = -100\)[/tex].
- The discriminant is [tex]\(b^2 - 4ac = 20^2 - 4(1)(-100) = 400 + 400 = 800\)[/tex], which is a positive number. Therefore, the quadratic can be factored.
Going back to the original form with [tex]\(x\)[/tex], this implies it can be factored further.
However, just a quick check shows that attempting to factor it over integers might not lead to integer factors directly.
Therefore, the expression was incorrectly guessed from running the code, and it turns out, it should remain as is. In this context, based on integer factorization, none of these polynomials simplify in that straightforward manner as was thought from above inspection, leaving this particular polynomial as in suspect belief, however since we shall trust the evaluation procedures it emerges.
Given this scenario in the steps above, option D appears not to factor with simple integer-based factorization despite the initial indications when evaluated manually like was suspected in above steps.
So, the prime polynomial is:
- D. [tex]\(x^4 + 20x^2 - 100\)[/tex]
This expression isn't obviously factorable using integer simple techniques, prompting verification with theoretical understanding.
A. [tex]\(x^3 - 27y^6\)[/tex]
This expression can be factored because it is a difference of cubes. The formula for the difference of cubes is [tex]\(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)[/tex]. Here, [tex]\(x^3\)[/tex] is [tex]\(a^3\)[/tex] and [tex]\( (3y^2)^3\)[/tex] is [tex]\(b^3\)[/tex]. So, this can be factored and is not a prime polynomial.
B. [tex]\(3x^2 + 18y\)[/tex]
This expression can be factored by taking out the greatest common factor (GCF). The GCF in this case is 3, so it can be rewritten as [tex]\(3(x^2 + 6y)\)[/tex]. Since it can be factored, it is not a prime polynomial.
C. [tex]\(10x^4 - 5x^3 + 70x^2 + 3x\)[/tex]
We can factor out an [tex]\(x\)[/tex] as a common factor, resulting in [tex]\(x(10x^3 - 5x^2 + 70x + 3)\)[/tex]. Since it can be factored further, it is not a prime polynomial.
D. [tex]\(x^4 + 20x^2 - 100\)[/tex]
This polynomial is a bit more challenging. We can try to factor this as a quadratic in form. Let [tex]\(z = x^2\)[/tex]. Then, the expression becomes [tex]\(z^2 + 20z - 100\)[/tex], which is quadratic in [tex]\(z\)[/tex].
Let's use the quadratic formula [tex]\(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] to check the roots:
- Here, [tex]\(a = 1\)[/tex], [tex]\(b = 20\)[/tex], and [tex]\(c = -100\)[/tex].
- The discriminant is [tex]\(b^2 - 4ac = 20^2 - 4(1)(-100) = 400 + 400 = 800\)[/tex], which is a positive number. Therefore, the quadratic can be factored.
Going back to the original form with [tex]\(x\)[/tex], this implies it can be factored further.
However, just a quick check shows that attempting to factor it over integers might not lead to integer factors directly.
Therefore, the expression was incorrectly guessed from running the code, and it turns out, it should remain as is. In this context, based on integer factorization, none of these polynomials simplify in that straightforward manner as was thought from above inspection, leaving this particular polynomial as in suspect belief, however since we shall trust the evaluation procedures it emerges.
Given this scenario in the steps above, option D appears not to factor with simple integer-based factorization despite the initial indications when evaluated manually like was suspected in above steps.
So, the prime polynomial is:
- D. [tex]\(x^4 + 20x^2 - 100\)[/tex]
This expression isn't obviously factorable using integer simple techniques, prompting verification with theoretical understanding.
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