Question

which polynomial is prime?
\\( x^4 + 3x^2 - x^2 - 3 \\)
\\( x^4 - 3x^2 - x^2 + 3 \\)
\\( 3x^2 + x - 6x - 2 \\)
\\( 3x^2 + x - 6x + 3 \\)

Explanation:

Step1: Recall Prime Polynomial Definition

A prime polynomial (irreducible polynomial) cannot be factored into non - constant polynomials with integer coefficients. We will try to factor each given polynomial.

Step2: Analyze \(x^{4}+3x^{2}-x^{2}-3\)

First, simplify the polynomial: \(x^{4}+(3x^{2}-x^{2}) - 3=x^{4}+2x^{2}-3\). Let \(y = x^{2}\), then the polynomial becomes \(y^{2}+2y - 3\).
Factor \(y^{2}+2y - 3\): We need two numbers that multiply to \(- 3\) and add up to \(2\). The numbers are \(3\) and \(-1\). So \(y^{2}+2y - 3=(y + 3)(y - 1)\). Substituting back \(y=x^{2}\), we get \((x^{2}+3)(x^{2}-1)=(x^{2}+3)(x + 1)(x - 1)\). So this polynomial is not prime.

Step3: Analyze \(x^{4}-3x^{2}-x^{2}+3\)

Simplify the polynomial: \(x^{4}+(-3x^{2}-x^{2})+3=x^{4}-4x^{2}+3\). Let \(y = x^{2}\), then the polynomial becomes \(y^{2}-4y + 3\).
Factor \(y^{2}-4y + 3\): We need two numbers that multiply to \(3\) and add up to \(-4\). The numbers are \(-1\) and \(-3\). So \(y^{2}-4y + 3=(y - 1)(y - 3)\). Substituting back \(y = x^{2}\), we get \((x^{2}-1)(x^{2}-3)=(x + 1)(x - 1)(x^{2}-3)\). So this polynomial is not prime.

Step4: Analyze \(3x^{2}+x - 6x - 2\)

Group the terms: \((3x^{2}+x)+(-6x - 2)=x(3x + 1)-2(3x + 1)=(3x + 1)(x - 2)\). So this polynomial is not prime.

Step5: Analyze \(3x^{2}+x - 6x + 3\)

Simplify the middle terms: \(3x^{2}-5x + 3\).
To check if it can be factored, we use the discriminant of the quadratic formula for \(ax^{2}+bx + c\) (here \(a = 3\), \(b=-5\), \(c = 3\)). The discriminant \(\Delta=b^{2}-4ac=(-5)^{2}-4\times3\times3=25 - 36=-11\). Since the discriminant is negative and we are factoring over the integers, and also, we can try to find two numbers \(m\) and \(n\) such that \(m\times n=3\times3 = 9\) and \(m + n=-5\). But there are no two integers \(m\) and \(n\) that satisfy this. Also, we can't group - factor it. Let's try to see if we can factor it as \((ax + b)(cx + d)=acx^{2}+(ad+bc)x+bd\). We have \(ac = 3\), \(bd = 3\), and \(ad + bc=-5\). The possible values for \(a = 3\), \(c = 1\) (or vice - versa) and \(b = 3\), \(d = 1\) (or vice - versa) give \(ad+bc=3\times1+3\times1 = 6\) or \(3\times3+1\times1=10\) or other combinations don't give \(-5\). So \(3x^{2}-5x + 3\) (after simplifying \(3x^{2}+x - 6x + 3\)) is a prime polynomial.

Answer:

\(3x^{2}+x - 6x + 3\) (the fourth polynomial in the list)