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Answer :
Sure! Let's figure out which expression is a prime polynomial by analyzing each option:
A prime polynomial is one that cannot be factored into polynomials of lower degree with integer coefficients.
Let's look at each option individually:
A. [tex]\( x^3 - 27y^6 \)[/tex]
This expression closely resembles the difference of cubes: [tex]\( a^3 - b^3 \)[/tex]. Here, [tex]\( x \)[/tex] is [tex]\( a \)[/tex] and [tex]\( (3y^2) \)[/tex] is [tex]\( b \)[/tex]. It can be factored as:
[tex]\[
x^3 - (3y^2)^3 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4)
\][/tex]
Since it can be factored, it is not a prime polynomial.
B. [tex]\( x^4 + 20x^2 - 100 \)[/tex]
Let's try factoring this polynomial. By substituting [tex]\( z = x^2 \)[/tex], we transform it to:
[tex]\[
z^2 + 20z - 100
\][/tex]
We need to check if it can be factored into the product of two binomials:
The quadratic formula tells us the roots are:
[tex]\[
z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-20 \pm \sqrt{20^2 + 400}}{2} = \frac{-20 \pm \sqrt{800}}{2}
\][/tex]
The solutions are not integers, thus the quadratic cannot be factored nicely, indicating that the original polynomial is prime.
C. [tex]\( 10x^4 - 5x^3 + 70x^2 + 3x \)[/tex]
We can factor out an [tex]\( x \)[/tex]:
[tex]\[
x(10x^3 - 5x^2 + 70x + 3)
\][/tex]
Since it is factored, it is not a prime polynomial.
D. [tex]\( 3x^2 + 18y \)[/tex]
This polynomial can be factored by taking out the greatest common factor, which is 3:
[tex]\[
3(x^2 + 6y)
\][/tex]
Since it can be factored, it is not a prime polynomial.
Based on this, the expression that is a prime polynomial is:
B. [tex]\( x^4 + 20x^2 - 100 \)[/tex]
A prime polynomial is one that cannot be factored into polynomials of lower degree with integer coefficients.
Let's look at each option individually:
A. [tex]\( x^3 - 27y^6 \)[/tex]
This expression closely resembles the difference of cubes: [tex]\( a^3 - b^3 \)[/tex]. Here, [tex]\( x \)[/tex] is [tex]\( a \)[/tex] and [tex]\( (3y^2) \)[/tex] is [tex]\( b \)[/tex]. It can be factored as:
[tex]\[
x^3 - (3y^2)^3 = (x - 3y^2)(x^2 + 3xy^2 + 9y^4)
\][/tex]
Since it can be factored, it is not a prime polynomial.
B. [tex]\( x^4 + 20x^2 - 100 \)[/tex]
Let's try factoring this polynomial. By substituting [tex]\( z = x^2 \)[/tex], we transform it to:
[tex]\[
z^2 + 20z - 100
\][/tex]
We need to check if it can be factored into the product of two binomials:
The quadratic formula tells us the roots are:
[tex]\[
z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-20 \pm \sqrt{20^2 + 400}}{2} = \frac{-20 \pm \sqrt{800}}{2}
\][/tex]
The solutions are not integers, thus the quadratic cannot be factored nicely, indicating that the original polynomial is prime.
C. [tex]\( 10x^4 - 5x^3 + 70x^2 + 3x \)[/tex]
We can factor out an [tex]\( x \)[/tex]:
[tex]\[
x(10x^3 - 5x^2 + 70x + 3)
\][/tex]
Since it is factored, it is not a prime polynomial.
D. [tex]\( 3x^2 + 18y \)[/tex]
This polynomial can be factored by taking out the greatest common factor, which is 3:
[tex]\[
3(x^2 + 6y)
\][/tex]
Since it can be factored, it is not a prime polynomial.
Based on this, the expression that is a prime polynomial is:
B. [tex]\( x^4 + 20x^2 - 100 \)[/tex]
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