Question

select the correct answer.
which expression is a prime polynomial?
a. ( x^4 + 20x^2 - 100 )
b. ( 10x^4 - 5x^3 + 70x^2 + 3x )
c. ( 3x^2 + 18y )
d. ( x^3 - 27y^3 )

Explanation:

Step1: Recall Prime Polynomial Definition

A prime polynomial (or irreducible polynomial) cannot be factored into non - constant polynomials with integer coefficients (for polynomials with integer coefficients, which is the common case here). We will check each option for factorability.

Step2: Analyze Option A: \(x^{4}+20x^{2}-100\)

Let \(u = x^{2}\), then the polynomial becomes \(u^{2}+20u - 100\). Using the quadratic formula \(u=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(au^{2}+bu + c\), here \(a = 1\), \(b=20\), \(c=- 100\). \(\Delta=b^{2}-4ac=20^{2}-4\times1\times(- 100)=400 + 400=800\), and \(u=\frac{-20\pm\sqrt{800}}{2}=\frac{-20\pm20\sqrt{2}}{2}=-10\pm10\sqrt{2}\). But we can also try to factor it as a quadratic in \(x^{2}\). Wait, actually, \(x^{4}+20x^{2}-100\) can be factored? Wait, no, maybe I made a mistake. Wait, let's try to see if it's a quadratic in \(x^{2}\). The discriminant of \(x^{4}+20x^{2}-100\) (treating \(x^{2}\) as a variable) is \(20^{2}-4\times1\times(-100)=400 + 400 = 800\), which is not a perfect square. Wait, but maybe it can be factored in another way? Wait, no, actually, let's check the other options first.

Step3: Analyze Option B: \(10x^{4}-5x^{3}+70x^{2}+3x\)

We can factor out an \(x\) from each term: \(x(10x^{3}-5x^{2}+70x + 3)\). Since we can factor out a non - constant factor (\(x\)), this polynomial is not prime.

Step4: Analyze Option C: \(3x^{2}+18y\)

We can factor out a 3: \(3(x^{2}+6y)\). Since we can factor out a non - constant factor (3), this polynomial is not prime.

Step5: Analyze Option D: \(x^{3}-27y^{3}\)

This is a difference of cubes. The formula for \(a^{3}-b^{3}=(a - b)(a^{2}+ab + b^{2})\). Here \(a=x\) and \(b = 3y\), so \(x^{3}-27y^{3}=(x - 3y)(x^{2}+3xy+9y^{2})\). So it can be factored.

Wait, but earlier analysis of Option A: Wait, maybe I made a mistake. Let's re - check Option A. \(x^{4}+20x^{2}-100\). Let's try to see if it can be factored. Suppose it factors as \((x^{2}+ax + b)(x^{2}+cx + d)=x^{4}+(a + c)x^{3}+(ac + b + d)x^{2}+(ad+bc)x+bd\). Comparing coefficients: \(a + c = 0\) (since there is no \(x^{3}\) term), \(bd=-100\), \(ad + bc = 0\) (since there is no \(x\) term), and \(ac + b + d = 20\). From \(a + c=0\), we have \(c=-a\). Then \(ad+bc=ad - ab=a(d - b)=0\). So either \(a = 0\) or \(d = b\). If \(a = 0\), then \(c = 0\), and \(ac + b + d=b + d = 20\), and \(bd=-100\). Solving \(b + d=20\) and \(bd=-100\), the quadratic equation is \(t^{2}-20t - 100 = 0\), discriminant \(400+400 = 800\), not a perfect square, so no integer solutions. If \(d = b\), then \(bd=b^{2}=-100\), which has no real (let alone integer) solutions. So \(x^{4}+20x^{2}-100\) cannot be factored into polynomials with integer coefficients (we tried factoring as a product of two quadratics with integer coefficients and it didn't work, and we saw that factoring out a common factor is not possible). Wait, but earlier when we considered Option D, we saw that it factors as a difference of cubes. Option B factors out an \(x\), Option C factors out a 3. So the only one that cannot be factored (prime) is Option A? Wait, no, wait, let's re - check Option A. Wait, the problem says "prime polynomial". A prime polynomial in one variable (or with multiple variables) over the integers is a polynomial that cannot be factored into non - constant polynomials with integer coefficients.

Wait, Option A: \(x^{4}+20x^{2}-100\). Let's check the degree. It's a degree 4 polynomial. We tried factoring as a product of two quadratics with integer coefficients and it's not possible (since the discriminant of the quadratic in \(x^{2}\) is not a perfect square, so it doesn't factor into linear factors in \(x^{2}\) with integer coefficients, and factoring into quadratics with integer coefficients also doesn't work as we saw from the system of equations). Option B: factors into \(x\) times a cubic, so not prime. Option C: factors into 3 times a binomial, so not prime. Option D: factors as a difference of cubes, so not prime. So the correct answer is A.

Answer:

A. \(x^{4}+20x^{2}-100\)