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

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.

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)