Determine if Prime 6y^2-21y-45

$6 y^{2} - 21 y - 45$

Step 1

Find the discriminant for $6 y^{2} - 21 y - 45 = 0$ . In this case, $b^{2} - 4 a c = 1521$ .

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 $c$ .

${(- 21)}^{2} - 4 (6 \cdot - 45)$

Evaluate the result to find the discriminant.

Simplify each term.

Raise $- 21$ to the power of $2$ .

$441 - 4 (6 \cdot - 45)$

Multiply $- 4 (6 \cdot - 45)$ .

Multiply $6$ by $- 45$ .

$441 - 4 \cdot - 270$

Multiply $- 4$ by $- 270$ .

$441 + 1080$

Add $441$ and $1080$ .

$1521$

Step 2

A perfect square number is an integer that is the square of another integer. $\sqrt{1521} = 39$ , which is an integer number.

$\sqrt{1521} = 39$

Step 3

Since $1521$ is the square of $39$ , it is a perfect square number.

$1521$ is a perfect square number

Step 4

The polynomial $6 y^{2} - 21 y - 45$ is not prime because the discriminant is a perfect square number.

Not prime