Answer:
1. The given equation does not have root at x = - 3.
2. (x - 3) (x + 1) (x² + 4) = 0
This curve will intersect twice the x-axis.
Step-by-step explanation:
1. If the right hand side of the equation [tex]x^{3} +3x^{2} +x+1 =0[/tex] ......... (1) becomes same as the left hand side by putting x = - 3, then only we can conclude that x = - 3 is a root of the equation.
But in this case [tex](-3)^{3} + 3(-3)^{2}+(-3) +1 \neq 0[/tex].
Therefore, x = - 3 is not a root of the above equation (1).
2. The equation of the polynomial with roots 3, -1, 2i, and -2i is
(x - 3) (x + 1) (x - 2i) (x + 2i) = 0
⇒ (x - 3) (x + 1) (x² + 4) = 0 (Answer)
Therefore, the graph of the above curve will intersect twice the x-axis in a real coordinate plane, at x = 3 and at x = -1. (Answer)
