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Answer :
To determine which expression is a prime polynomial, we need to find the polynomial that cannot be factored into polynomials of lower degree with integer coefficients. Let's examine each option:
A. [tex]\( x^3 - 27y^6 \)[/tex]
- This is a difference of cubes because [tex]\( x^3 - (3y^2)^3 \)[/tex].
- The difference of cubes formula is [tex]\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)[/tex].
- Substituting in, we get [tex]\( (x - 3y^2)(x^2 + 3xy^2 + 9y^4) \)[/tex].
- Since it can be factored, this is not a prime polynomial.
B. [tex]\( 3x^2 + 18y \)[/tex]
- This expression has a greatest common factor (GCF) of 3.
- Factoring out the GCF gives [tex]\( 3(x^2 + 6y) \)[/tex].
- Since it can be factored, this is not a prime polynomial.
C. [tex]\( 10x^4 - 5x^3 + 70x^2 + 3x \)[/tex]
- This expression can be examined for any common factors or factor patterns.
- Factoring out any common factors or seeing if it can be simplified doesn't yield a factorable expression with integer coefficients.
- However, it turns out this polynomial actually isn't considered a prime polynomial either.
D. [tex]\( x^4 + 20x^2 - 100 \)[/tex]
- This can be treated as a quadratic in terms of [tex]\( x^2 \)[/tex]: [tex]\( (x^2)^2 + 20(x^2) - 100 \)[/tex].
- Testing if it can be factored: let [tex]\( u = x^2 \)[/tex], then [tex]\( u^2 + 20u - 100 \)[/tex].
- The discriminant of this quadratic, [tex]\( b^2 - 4ac \)[/tex], is more than zero (it is 900 - a perfect square), meaning it can be factored into two binomials.
- Factoring results in [tex]\( (x^2 + 10)^2 - 10\)[/tex], showing it's reducible.
Given these observations, the results tell us:
- Expression B, [tex]\( 3x^2 + 18y \)[/tex], is considered to be a prime polynomial because it is not reducible into smaller degree polynomials with integer coefficients once we factor out the GCF.
In conclusion, option B is the prime polynomial.
A. [tex]\( x^3 - 27y^6 \)[/tex]
- This is a difference of cubes because [tex]\( x^3 - (3y^2)^3 \)[/tex].
- The difference of cubes formula is [tex]\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)[/tex].
- Substituting in, we get [tex]\( (x - 3y^2)(x^2 + 3xy^2 + 9y^4) \)[/tex].
- Since it can be factored, this is not a prime polynomial.
B. [tex]\( 3x^2 + 18y \)[/tex]
- This expression has a greatest common factor (GCF) of 3.
- Factoring out the GCF gives [tex]\( 3(x^2 + 6y) \)[/tex].
- Since it can be factored, this is not a prime polynomial.
C. [tex]\( 10x^4 - 5x^3 + 70x^2 + 3x \)[/tex]
- This expression can be examined for any common factors or factor patterns.
- Factoring out any common factors or seeing if it can be simplified doesn't yield a factorable expression with integer coefficients.
- However, it turns out this polynomial actually isn't considered a prime polynomial either.
D. [tex]\( x^4 + 20x^2 - 100 \)[/tex]
- This can be treated as a quadratic in terms of [tex]\( x^2 \)[/tex]: [tex]\( (x^2)^2 + 20(x^2) - 100 \)[/tex].
- Testing if it can be factored: let [tex]\( u = x^2 \)[/tex], then [tex]\( u^2 + 20u - 100 \)[/tex].
- The discriminant of this quadratic, [tex]\( b^2 - 4ac \)[/tex], is more than zero (it is 900 - a perfect square), meaning it can be factored into two binomials.
- Factoring results in [tex]\( (x^2 + 10)^2 - 10\)[/tex], showing it's reducible.
Given these observations, the results tell us:
- Expression B, [tex]\( 3x^2 + 18y \)[/tex], is considered to be a prime polynomial because it is not reducible into smaller degree polynomials with integer coefficients once we factor out the GCF.
In conclusion, option B is the prime polynomial.
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