Determine if Prime v^2+3v-4

$v^{2} + 3 v - 4$

Step 1

Find the discriminant for $v^{2} + 3 v - 4 = 0$ . In this case, $b^{2} - 4 a c = 25$ .

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 $c$ .

$3^{2} - 4 (1 \cdot - 4)$

Evaluate the result to find the discriminant.

Simplify each term.

Raise $3$ to the power of $2$ .

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

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

Multiply $- 4$ by $1$ .

$9 - 4 \cdot - 4$

Multiply $- 4$ by $- 4$ .

$9 + 16$

Add $9$ and $16$ .

$25$

Step 2

A perfect square number is an integer that is the square of another integer. $\sqrt{25} = 5$ , which is an integer number.

$\sqrt{25} = 5$

Step 3

Since $25$ is the square of $5$ , it is a perfect square number.

$25$ is a perfect square number

Step 4

The polynomial $v^{2} + 3 v - 4$ is not prime because the discriminant is a perfect square number.

Not prime