Examples

x^{2} - 4 x + 4

$x^{2} - 4 x + 4$

Step 1

Find the discriminant for

x^{2} - 4 x + 4 = 0

$x^{2} - 4 x + 4 = 0$ . In this case,

b^{2} - 4 a c = 0

$b^{2} - 4 a c = 0$ .

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Step 1.1

The discriminant of a quadratic is the expression inside the radical of the quadratic formula.

b^{2} - 4 (a c)

$b^{2} - 4 (a c)$

Step 1.2

Substitute in the values of

a

$a$ ,

b

$b$ , and

c

$c$ .

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

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

Step 1.3

Evaluate the result to find the discriminant.

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Step 1.3.1

Simplify each term.

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Step 1.3.1.1

Raise

- 4

$- 4$ to the power of

2

$2$ .

16 - 4 (1 \cdot 4)

$16 - 4 (1 \cdot 4)$

Step 1.3.1.2

Multiply

- 4 (1 \cdot 4)

$- 4 (1 \cdot 4)$ .

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Step 1.3.1.2.1

Multiply

4

$4$ by

1

$1$ .

16 - 4 \cdot 4

$16 - 4 \cdot 4$

Step 1.3.1.2.2

Multiply

- 4

$- 4$ by

4

$4$ .

16 - 16

$16 - 16$

16 - 16

$16 - 16$

16 - 16

$16 - 16$

Step 1.3.2

Subtract

16

$16$ from

16

$16$ .

0

$0$

0

$0$

0

$0$

Step 2

A perfect square number is an integer that is the square of another integer.

\sqrt{0} = 0

$\sqrt{0} = 0$ , which is an integer number.

\sqrt{0} = 0

$\sqrt{0} = 0$

Step 3

Since

0

$0$ is the square of

0

$0$ , it is a perfect square number.

0

$0$ is a perfect square number

Step 4

The polynomial

x^{2} - 4 x + 4

$x^{2} - 4 x + 4$ is not prime because the discriminant is a perfect square number.

Not prime

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