which polynomial is prime?\n$x^{2}-36$\n$x^{2}+6$\n$x^{2}-7x + 12$\n$x^{2}-x - 20$

Explanation:

Step1: Factorize $x^{2}-36$

Using the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$, where $a=x$ and $b = 6$. So $x^{2}-36=(x + 6)(x - 6)$.

Step2: Analyze $x^{2}+6$

There are no two real - valued binomials $(ax + b)(cx + d)$ such that $(ax + b)(cx + d)=x^{2}+6$. In the real number system, it cannot be factored into two non - constant polynomials with real coefficients.

Step3: Factorize $x^{2}-7x + 12$

We need to find two numbers that multiply to $12$ and add up to $-7$. The numbers are $-3$ and $-4$. So $x^{2}-7x + 12=(x - 3)(x - 4)$.

Step4: Factorize $x^{2}-x - 20$

We need to find two numbers that multiply to $-20$ and add up to $-1$. The numbers are $-5$ and $4$. So $x^{2}-x - 20=(x - 5)(x+4)$.

Answer:

$x^{2}+6$