Question

What are the solutions of the equation x^6 + 6x^3 + 5 = 0? Use factoring to solve

A) x = ^3√5 and x = 1

B) x = -^3√5 and x = -1

C) x = ^3√5 and x = -1

D) x = -^3√5 and x = 1

Answer 1

`B)\quad x=-\sqrt[3]{5}\quad\text{and}\quad x=-1`

Explanation:

The equation is a quadratic in x³, so can be factored the same way the quadratic x² +6x +5 = 0 would be factored.

... (x³ +1)(x³ +5) = 0

The solutions are the values of x that make the factors be zero. That is, the expression resolves to two cubic equations. (Each of those has 1 real root and 2 complex roots. We'll ignore the complex roots.)

For the first factor:

... x³ +1 = 0

... x³ = -1

... x = ∛(-1) = -1 . . . . . real root (complex roots ignored)

For the second factor:

... x³ +5 = 0

... x³ = -5

... x = ∛(-5) = -∛5 . . . . . real root (complex roots ignored)

The real solutions to the equation are x = -∛5 and x = -1.