Answer :
Sure! Let's determine which expression is a prime polynomial. A polynomial is considered prime if it is not factorable using real or complex numbers (other than the polynomial itself and 1).
Let's examine each option:
A. [tex]\(x^4 + 20x^2 - 100\)[/tex]
This is a quadratic in form (by considering [tex]\( x^2 = y \)[/tex], it can be rewritten as [tex]\( y^2 + 20y - 100 \)[/tex]). We can see if it factors by attempting some common factoring methods, like searching for roots or checking for simple factors. However, a quick check or use of factoring techniques shows it cannot be easily factored over the integers. Therefore, it might be prime.
B. [tex]\(3x^2 + 18y\)[/tex]
This expression can actually be factored by taking out a common factor of 3:
[tex]\[ 3x^2 + 18y = 3(x^2 + 6y) \][/tex]
This means the expression is not prime because it can be factored further.
C. [tex]\(10x^4 - 5x^3 + 70x^2 + 3x\)[/tex]
We can factor this expression by taking out the greatest common factor, which is [tex]\(x\)[/tex]:
[tex]\[ 10x^4 - 5x^3 + 70x^2 + 3x = x(10x^3 - 5x^2 + 70x + 3) \][/tex]
Since the expression can be factored with [tex]\(x\)[/tex] as a common term, it is not prime.
D. [tex]\(x^3 - 27y^6\)[/tex]
This expression is actually a difference of cubes. It can be factored using the formula for factoring the difference of cubes:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
Here, [tex]\(a = x\)[/tex] and [tex]\(b = 3y^2\)[/tex], so
[tex]\[ x^3 - (3y^2)^3 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4) \][/tex]
Since it can be factored this way, it is not prime.
Conclusion:
The expression that turns out to be non-factorable (or prime) among the given options is likely option A, [tex]\(x^4 + 20x^2 - 100\)[/tex], as it doesn't factor neatly into simpler polynomials over the integers or reals.
Let's examine each option:
A. [tex]\(x^4 + 20x^2 - 100\)[/tex]
This is a quadratic in form (by considering [tex]\( x^2 = y \)[/tex], it can be rewritten as [tex]\( y^2 + 20y - 100 \)[/tex]). We can see if it factors by attempting some common factoring methods, like searching for roots or checking for simple factors. However, a quick check or use of factoring techniques shows it cannot be easily factored over the integers. Therefore, it might be prime.
B. [tex]\(3x^2 + 18y\)[/tex]
This expression can actually be factored by taking out a common factor of 3:
[tex]\[ 3x^2 + 18y = 3(x^2 + 6y) \][/tex]
This means the expression is not prime because it can be factored further.
C. [tex]\(10x^4 - 5x^3 + 70x^2 + 3x\)[/tex]
We can factor this expression by taking out the greatest common factor, which is [tex]\(x\)[/tex]:
[tex]\[ 10x^4 - 5x^3 + 70x^2 + 3x = x(10x^3 - 5x^2 + 70x + 3) \][/tex]
Since the expression can be factored with [tex]\(x\)[/tex] as a common term, it is not prime.
D. [tex]\(x^3 - 27y^6\)[/tex]
This expression is actually a difference of cubes. It can be factored using the formula for factoring the difference of cubes:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
Here, [tex]\(a = x\)[/tex] and [tex]\(b = 3y^2\)[/tex], so
[tex]\[ x^3 - (3y^2)^3 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4) \][/tex]
Since it can be factored this way, it is not prime.
Conclusion:
The expression that turns out to be non-factorable (or prime) among the given options is likely option A, [tex]\(x^4 + 20x^2 - 100\)[/tex], as it doesn't factor neatly into simpler polynomials over the integers or reals.
