Determine if Prime x^2+12x+28

$x^{2} + 12 x + 28$

Find the discriminant for $x^{2} + 12 x + 28 = 0$ . In this case, $b^{2} - 4 a c = 32$ .

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 $c$ .

$12^{2} - 4 (1 \cdot 28)$

Evaluate the result to find the discriminant.

Simplify each term.

Raise $12$ to the power of $2$ .

$144 - 4 (1 \cdot 28)$

Multiply $28$ by $1$ .

$144 - 4 \cdot 28$

Multiply $- 4$ by $28$ .

$144 - 112$

Subtract $112$ from $144$ .

$32$

A perfect square number is an integer that is the square of another integer. $\sqrt{32} \approx 5.65685424$ , which is not an integer number.

$\sqrt{32} \approx 5.65685424$

Since $32$ can\'t be the square of another integer, it is not a perfect square number.

The polynomial $x^{2} + 12 x + 28$ is prime because the discriminant is not a perfect square number.

Prime