Determine if Prime x^2-9

$x^{2} - 9$

Step 1

Find the discriminant for $x^{2} - 9 = 0$ . In this case, $b^{2} - 4 a c = 36$ .

The discriminant of a quadratic is the expression inside the radical of the quadratic formula.

$b^{2} - 4 (a c)$

Substitute in the values of $a$ , $b$ , and $c$ .

$0^{2} - 4 (1 \cdot - 9)$

Evaluate the result to find the discriminant.

Simplify each term.

Raising $0$ to any positive power yields $0$ .

$0 - 4 (1 \cdot - 9)$

Multiply $- 4 (1 \cdot - 9)$ .

Multiply $- 9$ by $1$ .

$0 - 4 \cdot - 9$

Multiply $- 4$ by $- 9$ .

$0 + 36$

Add $0$ and $36$ .

$36$

Step 2

A perfect square number is an integer that is the square of another integer. $\sqrt{36} = 6$ , which is an integer number.

$\sqrt{36} = 6$

Step 3

Since $36$ is the square of $6$ , it is a perfect square number.

$36$ is a perfect square number

Step 4

The polynomial $x^{2} - 9$ is not prime because the discriminant is a perfect square number.

Not prime