Question

From the polynomial p(x) = x² + x + 1, do we get p(x) = 0 for any x? What about p(x) = 1? And p(x) = −1?

Answer 1

p(x) = x² + x + 1

p(x) = 0

⇒ x² + x + 1 = 0

Comparing the above equation with general quadratic equation:

a = 1, b = 1 and c = 1

Discriminant of the above equation is:

b² − 4ac = (1)² − 4(1)(1) = 1 − 4 = −3

Since, the discriminant is negative, p(x) ≠ 0 for any x.

p(x) = 1

⇒ x² + x + 1 = 1

⇒ x² + x + 1 − 1 = 0

⇒ x² + x + 0 = 0

Comparing the above equation with general quadratic equation:

a = 1, b = 1 and c = 0

Discriminant of the above equation is:

b² − 4ac = (1)² − 4(1)(0)

= 1 − 0

= 1

∴ p(x) = 1 for x = −1 or x = 0.

Now, p(x) = −1

⇒ x² + x + 1 = −1

⇒ x² + x + 1 + 1 = 0

⇒ x² + x + 2 = 0

Comparing the above equation with general quadratic equation:

a = 1, b = 1 and c = 2

Discriminant of the above equation is:

b² − 4ac = (1)² − 4(1)(2)

= 1 − 8

= −7

Since the discriminant is negative, p(x) ≠ −1 for any x.