Determine if Prime x^2+10x+21

$x^{2} + 10 x + 21$

Step 1

Find the discriminant for $x^{2} + 10 x + 21 = 0$ . In this case, $b^{2} - 4 a c = 16$ .

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 $c$ .

$10^{2} - 4 (1 \cdot 21)$

Evaluate the result to find the discriminant.

Simplify each term.

Raise $10$ to the power of $2$ .

$100 - 4 (1 \cdot 21)$

Multiply $- 4 (1 \cdot 21)$ .

Multiply $21$ by $1$ .

$100 - 4 \cdot 21$

Multiply $- 4$ by $21$ .

$100 - 84$

Subtract $84$ from $100$ .

$16$

Step 2

A perfect square number is an integer that is the square of another integer. $\sqrt{16} = 4$ , which is an integer number.

$\sqrt{16} = 4$

Step 3

Since $16$ is the square of $4$ , it is a perfect square number.

$16$ is a perfect square number

Step 4

The polynomial $x^{2} + 10 x + 21$ is not prime because the discriminant is a perfect square number.

Not prime