Answer :
To determine which of the given polynomial expressions is a prime polynomial, we need to figure out which one cannot be factored into polynomials of lower degree over the set of integers.
The expressions given in the question are:
A. [tex]\(10x^4 - 5x^3 + 70x^2 + 3\)[/tex]
B. [tex]\(x^3 - 27y^6\)[/tex]
C. [tex]\(x^4 + 20x^2 - 100\)[/tex]
D. [tex]\(3x^2 + 18y\)[/tex]
Step-by-Step Solution:
1. Expression A: [tex]\(10x^4 - 5x^3 + 70x^2 + 3\)[/tex]
- We have to check if this polynomial can be factored. The polynomial appears to be quite complex, but let's consider whether there are any obvious ways to factor it. After checking possible factors, it turns out this polynomial is not easily factorable. This indicates it might be a candidate for being prime.
2. Expression B: [tex]\(x^3 - 27y^6\)[/tex]
- We recognize that this expression is a difference of cubes: [tex]\(x^3 - (3y^2)^3\)[/tex].
- Using the factorization formula for the difference of cubes, [tex]\(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)[/tex], we can factor this as:
[tex]\[
x^3 - 27y^6 = x^3 - (3y^2)^3 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4)
\][/tex]
- Since this polynomial can be factored, it is not a prime polynomial.
3. Expression C: [tex]\(x^4 + 20x^2 - 100\)[/tex]
- We can attempt to factor this expression by treating it as a quadratic in terms of [tex]\(x^2\)[/tex]:
[tex]\[
t = x^2 \quad \text{so} \quad t^2 + 20t - 100
\][/tex]
- Solving this quadratic using the quadratic formula [tex]\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] yields:
[tex]\[
t = \frac{-20 \pm \sqrt{400 + 400}}{2} = \frac{-20 \pm \sqrt{800}}{2} = -10 \pm 10\sqrt{2}
\][/tex]
- The polynomial can thus be written as:
[tex]\[
x^2 = -10 \pm 10\sqrt{2}
\][/tex]
- This indicates that the polynomial can actually be factored further (factorization involving square roots), hence it isn't prime.
4. Expression D: [tex]\(3x^2 + 18y\)[/tex]
- We see that both terms have a common factor of [tex]\(3\)[/tex]:
[tex]\[
3x^2 + 18y = 3(x^2 + 6y)
\][/tex]
- Since this polynomial can be factored, it is not a prime polynomial.
After examining all the expressions, the one that cannot be easily factored is:
A. [tex]\(10x^4 - 5x^3 + 70x^2 + 3\)[/tex]
Therefore, the correct answer is:
A. [tex]\(10x^4 - 5x^3 + 70x^2 + 3\)[/tex]
The expressions given in the question are:
A. [tex]\(10x^4 - 5x^3 + 70x^2 + 3\)[/tex]
B. [tex]\(x^3 - 27y^6\)[/tex]
C. [tex]\(x^4 + 20x^2 - 100\)[/tex]
D. [tex]\(3x^2 + 18y\)[/tex]
Step-by-Step Solution:
1. Expression A: [tex]\(10x^4 - 5x^3 + 70x^2 + 3\)[/tex]
- We have to check if this polynomial can be factored. The polynomial appears to be quite complex, but let's consider whether there are any obvious ways to factor it. After checking possible factors, it turns out this polynomial is not easily factorable. This indicates it might be a candidate for being prime.
2. Expression B: [tex]\(x^3 - 27y^6\)[/tex]
- We recognize that this expression is a difference of cubes: [tex]\(x^3 - (3y^2)^3\)[/tex].
- Using the factorization formula for the difference of cubes, [tex]\(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\)[/tex], we can factor this as:
[tex]\[
x^3 - 27y^6 = x^3 - (3y^2)^3 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4)
\][/tex]
- Since this polynomial can be factored, it is not a prime polynomial.
3. Expression C: [tex]\(x^4 + 20x^2 - 100\)[/tex]
- We can attempt to factor this expression by treating it as a quadratic in terms of [tex]\(x^2\)[/tex]:
[tex]\[
t = x^2 \quad \text{so} \quad t^2 + 20t - 100
\][/tex]
- Solving this quadratic using the quadratic formula [tex]\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex] yields:
[tex]\[
t = \frac{-20 \pm \sqrt{400 + 400}}{2} = \frac{-20 \pm \sqrt{800}}{2} = -10 \pm 10\sqrt{2}
\][/tex]
- The polynomial can thus be written as:
[tex]\[
x^2 = -10 \pm 10\sqrt{2}
\][/tex]
- This indicates that the polynomial can actually be factored further (factorization involving square roots), hence it isn't prime.
4. Expression D: [tex]\(3x^2 + 18y\)[/tex]
- We see that both terms have a common factor of [tex]\(3\)[/tex]:
[tex]\[
3x^2 + 18y = 3(x^2 + 6y)
\][/tex]
- Since this polynomial can be factored, it is not a prime polynomial.
After examining all the expressions, the one that cannot be easily factored is:
A. [tex]\(10x^4 - 5x^3 + 70x^2 + 3\)[/tex]
Therefore, the correct answer is:
A. [tex]\(10x^4 - 5x^3 + 70x^2 + 3\)[/tex]
