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Answer :
To determine which expression is a prime polynomial, we need to check each option and see if it can be factored into simpler polynomials with integer coefficients. If an expression can be factored, it is not a prime polynomial. Let's analyze each option step by step:
### A. [tex]\( x^3 - 27y^6 \)[/tex]
This expression can be factored using the difference of cubes formula [tex]\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)[/tex]:
- Here, [tex]\( x^3 - (3y^2)^3 \)[/tex] can be factored as:
[tex]\[ x^3 - 27y^6 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4) \][/tex]
Since it can be factored, it is not a prime polynomial.
### B. [tex]\( x^4 + 20x^2 - 100 \)[/tex]
This expression is a quadratic in terms of [tex]\( x^2 \)[/tex], meaning we can rewrite it as:
- Let [tex]\( u = x^2 \)[/tex], so the expression becomes [tex]\( u^2 + 20u - 100 \)[/tex].
We can use the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] to try to factor it:
[tex]\[ u = \frac{-20 \pm \sqrt{20^2 - 4 \cdot 1 \cdot (-100)}}{2 \cdot 1} \][/tex]
[tex]\[ u = \frac{-20 \pm \sqrt{400 + 400}}{2} \][/tex]
[tex]\[ u = \frac{-20 \pm \sqrt{800}}{2} \][/tex]
[tex]\[ u = \frac{-20 \pm 20\sqrt{2}}{2} \][/tex]
[tex]\[ u = -10 \pm 10\sqrt{2} \][/tex]
Since we have real solutions, it means the polynomial can be factored as:
[tex]\[ x^4 + 20x^2 - 100 = (x^2 - (-10 + 10\sqrt{2}))(x^2 - (-10 - 10\sqrt{2})) \][/tex]
Since it can be factored, it is not a prime polynomial.
### C. [tex]\( 3x^2 + 18y \)[/tex]
We can factor out the greatest common factor (GCF) of this expression:
- The GCF is [tex]\( 3 \)[/tex], so:
[tex]\[ 3x^2 + 18y = 3(x^2 + 6y) \][/tex]
Since it can be factored, it is not a prime polynomial.
### D. [tex]\( 10x^4 - 5x^3 + 70x^2 + 3x \)[/tex]
We can factor out the greatest common factor (GCF) of this expression:
- The GCF is [tex]\( x \)[/tex], so:
[tex]\[ 10x^4 - 5x^3 + 70x^2 + 3x = x(10x^3 - 5x^2 + 70x + 3) \][/tex]
Since it can be factored, it is not a prime polynomial.
Given the analysis above, none of the given polynomials are prime. Therefore, there is no correct answer among the provided options.
### A. [tex]\( x^3 - 27y^6 \)[/tex]
This expression can be factored using the difference of cubes formula [tex]\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)[/tex]:
- Here, [tex]\( x^3 - (3y^2)^3 \)[/tex] can be factored as:
[tex]\[ x^3 - 27y^6 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4) \][/tex]
Since it can be factored, it is not a prime polynomial.
### B. [tex]\( x^4 + 20x^2 - 100 \)[/tex]
This expression is a quadratic in terms of [tex]\( x^2 \)[/tex], meaning we can rewrite it as:
- Let [tex]\( u = x^2 \)[/tex], so the expression becomes [tex]\( u^2 + 20u - 100 \)[/tex].
We can use the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] to try to factor it:
[tex]\[ u = \frac{-20 \pm \sqrt{20^2 - 4 \cdot 1 \cdot (-100)}}{2 \cdot 1} \][/tex]
[tex]\[ u = \frac{-20 \pm \sqrt{400 + 400}}{2} \][/tex]
[tex]\[ u = \frac{-20 \pm \sqrt{800}}{2} \][/tex]
[tex]\[ u = \frac{-20 \pm 20\sqrt{2}}{2} \][/tex]
[tex]\[ u = -10 \pm 10\sqrt{2} \][/tex]
Since we have real solutions, it means the polynomial can be factored as:
[tex]\[ x^4 + 20x^2 - 100 = (x^2 - (-10 + 10\sqrt{2}))(x^2 - (-10 - 10\sqrt{2})) \][/tex]
Since it can be factored, it is not a prime polynomial.
### C. [tex]\( 3x^2 + 18y \)[/tex]
We can factor out the greatest common factor (GCF) of this expression:
- The GCF is [tex]\( 3 \)[/tex], so:
[tex]\[ 3x^2 + 18y = 3(x^2 + 6y) \][/tex]
Since it can be factored, it is not a prime polynomial.
### D. [tex]\( 10x^4 - 5x^3 + 70x^2 + 3x \)[/tex]
We can factor out the greatest common factor (GCF) of this expression:
- The GCF is [tex]\( x \)[/tex], so:
[tex]\[ 10x^4 - 5x^3 + 70x^2 + 3x = x(10x^3 - 5x^2 + 70x + 3) \][/tex]
Since it can be factored, it is not a prime polynomial.
Given the analysis above, none of the given polynomials are prime. Therefore, there is no correct answer among the provided options.
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