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Answer :
To determine if a polynomial is prime, we need to see if it can be factored into a product of two or more non-constant polynomials. A prime polynomial, or irreducible polynomial, cannot be factored further over its coefficient field.
Step-by-step analysis of the given options:
Option A: [tex]\( 3x^2 + 18y \)[/tex]
Let's factor this polynomial:
- We can factor out the greatest common divisor, which is 3:
[tex]\[ 3x^2 + 18y = 3(x^2 + 6y) \][/tex]
Since it can be factored into [tex]\( 3 \)[/tex] and [tex]\( (x^2 + 6y) \)[/tex], it is not a prime polynomial.
Option B: [tex]\( x^4 + 20x^2 - 100 \)[/tex]
Let's factor this polynomial:
- Notice this is a quadratic in terms of [tex]\( x^2 \)[/tex]:
[tex]\[ x^4 + 20x^2 - 100 \][/tex]
- Let [tex]\( u = x^2 \)[/tex]. Then it becomes a quadratic equation:
[tex]\[ u^2 + 20u - 100 \][/tex]
We can use the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]. However, factoring directly, we have:
[tex]\[ x^4 + 20x^2 - 100 = (x^2 + 10)^2 - (10)^2 = (x^2 + 20 + 10) (x^2 + 20 - 10) = (x^2 + 30)(x^2 + 10) \][/tex]
Since it can be factored into [tex]\((x^2 + 30)\)[/tex] and [tex]\((x^2 + 10)\)[/tex], it is not a prime polynomial.
Option C: [tex]\( 10x^4 - 5x^3 + 70x^2 + 3x \)[/tex]
Let's factor this polynomial:
- First, factor out the greatest common divisor, which is [tex]\( x \)[/tex]:
[tex]\[ 10x^4 - 5x^3 + 70x^2 + 3x = x(10x^3 - 5x^2 + 70x + 3) \][/tex]
Since it can be factored into [tex]\( x \)[/tex] and [tex]\( 10x^3 - 5x^2 + 70x + 3 \)[/tex], it is not a prime polynomial.
Option D: [tex]\( x^3 - 27y^6 \)[/tex]
Let's factor this polynomial:
- This is a difference of cubes:
[tex]\[ x^3 - (3y^2)^3 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4) \][/tex]
Since it can be factored into [tex]\( (x - 3y^2) \)[/tex] and [tex]\( (x^2 + 3xy^2 + 9y^4) \)[/tex], it is not a prime polynomial.
Given the analysis above, we see that none of the options [tex]\( A, B, C,\)[/tex] or [tex]\( D \)[/tex] are prime polynomials because they can all be factored into products of lower-degree polynomials.
Therefore, the correct answer is:
[tex]\[ \boxed{0} \][/tex]
indicating that no polynomial among the given options is a prime polynomial.
Step-by-step analysis of the given options:
Option A: [tex]\( 3x^2 + 18y \)[/tex]
Let's factor this polynomial:
- We can factor out the greatest common divisor, which is 3:
[tex]\[ 3x^2 + 18y = 3(x^2 + 6y) \][/tex]
Since it can be factored into [tex]\( 3 \)[/tex] and [tex]\( (x^2 + 6y) \)[/tex], it is not a prime polynomial.
Option B: [tex]\( x^4 + 20x^2 - 100 \)[/tex]
Let's factor this polynomial:
- Notice this is a quadratic in terms of [tex]\( x^2 \)[/tex]:
[tex]\[ x^4 + 20x^2 - 100 \][/tex]
- Let [tex]\( u = x^2 \)[/tex]. Then it becomes a quadratic equation:
[tex]\[ u^2 + 20u - 100 \][/tex]
We can use the quadratic formula [tex]\( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]. However, factoring directly, we have:
[tex]\[ x^4 + 20x^2 - 100 = (x^2 + 10)^2 - (10)^2 = (x^2 + 20 + 10) (x^2 + 20 - 10) = (x^2 + 30)(x^2 + 10) \][/tex]
Since it can be factored into [tex]\((x^2 + 30)\)[/tex] and [tex]\((x^2 + 10)\)[/tex], it is not a prime polynomial.
Option C: [tex]\( 10x^4 - 5x^3 + 70x^2 + 3x \)[/tex]
Let's factor this polynomial:
- First, factor out the greatest common divisor, which is [tex]\( x \)[/tex]:
[tex]\[ 10x^4 - 5x^3 + 70x^2 + 3x = x(10x^3 - 5x^2 + 70x + 3) \][/tex]
Since it can be factored into [tex]\( x \)[/tex] and [tex]\( 10x^3 - 5x^2 + 70x + 3 \)[/tex], it is not a prime polynomial.
Option D: [tex]\( x^3 - 27y^6 \)[/tex]
Let's factor this polynomial:
- This is a difference of cubes:
[tex]\[ x^3 - (3y^2)^3 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4) \][/tex]
Since it can be factored into [tex]\( (x - 3y^2) \)[/tex] and [tex]\( (x^2 + 3xy^2 + 9y^4) \)[/tex], it is not a prime polynomial.
Given the analysis above, we see that none of the options [tex]\( A, B, C,\)[/tex] or [tex]\( D \)[/tex] are prime polynomials because they can all be factored into products of lower-degree polynomials.
Therefore, the correct answer is:
[tex]\[ \boxed{0} \][/tex]
indicating that no polynomial among the given options is a prime polynomial.
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