which polynomial is prime?\n$x^{4}+3x^{2}-x^{2}-3$\n$x^{4}-3x^{2}-x^{2}+3$\n$3x^{2}+x - 6x - 2$\n$3x^{2}+x…

Explanation:

Step1: Simplify and factor first option

Combine like - terms: $x^{4}+3x^{2}-x^{2}-3=x^{4}+2x^{2}-3$. Let $y = x^{2}$, then it becomes $y^{2}+2y - 3=(y + 3)(y - 1)=(x^{2}+3)(x^{2}-1)=(x^{2}+3)(x + 1)(x - 1)$.

Step2: Simplify and factor second option

Combine like - terms: $x^{4}-3x^{2}-x^{2}+3=x^{4}-4x^{2}+3$. Let $y=x^{2}$, then $y^{2}-4y + 3=(y - 3)(y - 1)=(x^{2}-3)(x^{2}-1)=(x^{2}-3)(x + 1)(x - 1)$.

Step3: Factor third option

Group the terms: $3x^{2}+x-6x - 2=(3x^{2}+x)-(6x + 2)=x(3x + 1)-2(3x + 1)=(3x + 1)(x - 2)$.

Step4: Check fourth option

For the polynomial $3x^{2}+x-6x + 3=3x^{2}-5x + 3$. The discriminant of a quadratic polynomial $ax^{2}+bx + c$ is $\Delta=b^{2}-4ac$. Here, $a = 3$, $b=-5$, $c = 3$. Then $\Delta=(-5)^{2}-4\times3\times3=25 - 36=-11<0$. Since the discriminant is negative, it cannot be factored over the real numbers.

Answer:

$3x^{2}+x-6x + 3$