X squared minus 5 is not necessarily prime. For instance; if x=3, then x squared minus 5 is 4, and 4 is not prime.
Please consider this addition to the above answer. I opted to keep the original answer there for a reason...
Whether x^2-5 is prime depends not on the values of x that you plug into the expression, but rather on the primality of the expression itself. In other words, the expression's primality (primeness) is based on the presence or absence of RATIONAL ROOTS. x^2-5 can be factored into (x+√5)(x-√5), making x=±√5. √5 is IRRATIONAL, and since no root of the original expression is rational, the original expression is PRIME. It is ALWAYS PRIME.
For the sake of improved clarity, consider the following examples:
x^2-4 is NOT PRIME because it can be represented by (x+2)(x-2), making x=±2, both of which are RATIONAL ROOTS.
x^2-3 is PRIME because it can be represented by (x+√3)(x-√3), making x=±3, both of which are IRRATIONAL.
I hope this helps! Please ask for clarification if you need it, as well as for any other form of assistance... including other questions.
(x−6)(x+5)
well, i believe it goes like this (not certain) 25 is the only not prime number so... 3xcubed x squared -5 squared
x squared plus 5
It is: (x+5)(x-8) when factored
(x - 5)(x + 3)
(x - 5)(x + 4)
(x−6)(x+5)
well, i believe it goes like this (not certain) 25 is the only not prime number so... 3xcubed x squared -5 squared
x squared plus 5
It is: (x+5)(x-8) when factored
Nope! It's (x+ 1) (x-1). :)
It is: (x+5)(x-8) when factored
(x - 5)(x + 3)
2
No
It is impossible but if it were x squared plus 2x minus 15 the equation would be (x+5) (x-3) with x being equal to either -5 or 3. If the original problem was x squared minus 2x minus 15 the equation would be (x-5)(x+3) and x would be equal to either 5 or -3
(x - 5)(x^2 + 1)