Determine if Prime v^2+4v+2

$v^{2} + 4 v + 2$

Step 1

Find the discriminant for $v^{2} + 4 v + 2 = 0$ . In this case, $b^{2} - 4 a c = 8$ .

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 $c$ .

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

Evaluate the result to find the discriminant.

Simplify each term.

Raise $4$ to the power of $2$ .

$16 - 4 (1 \cdot 2)$

Multiply $- 4 (1 \cdot 2)$ .

Multiply $2$ by $1$ .

$16 - 4 \cdot 2$

Multiply $- 4$ by $2$ .

$16 - 8$

Subtract $8$ from $16$ .

$8$

Step 2

A perfect square number is an integer that is the square of another integer. $\sqrt{8} \approx 2.82842712$ , which is not an integer number.

$\sqrt{8} \approx 2.82842712$

Step 3

Since $8$ can\'t be the square of another integer, it is not a perfect square number.

Step 4

The polynomial $v^{2} + 4 v + 2$ is prime because the discriminant is not a perfect square number.

Prime