The polynomial equation x(x² + 4)(x² - x - 6) = 0 has how many real roots?

Solution:

Given the polynomial:

x(x² + 4)(x² - x - 6 ) = 0

Equating each term to zero

x = 0 which is a real root

x² + 4 = 0

(x + 2i)(x - 2i) = 0

x = 2i,-2i which are two imaginary roots

(x² - x - 6) =0

x² - 3x + 2x - 6 = 0

x(x - 3) + 2(x - 3) = 0

(x + 2)(x - 3) = 0

x = -2,3 which are two real roots

Hence, the equation has 3 real roots.

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The polynomial equation x(x² + 4)(x² - x - 6) = 0 has how many real roots?

Summary:

The polynomial equation x(x² + 4)(x² - x - 6) = 0 has 3 real roots.