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 the first polynomial

Combine like - terms in (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 the second polynomial

Combine like - terms in (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 the third polynomial

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

Step4: Try to factor the fourth polynomial

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

Answer:

(3x^{2}+x-6x + 3)