To solve the equation \( 6(x-3)(x^2+4)(x+1)=0 \), we need to find the values of \( x \) that make each factor zero. Let's break it down step-by-step:
1. Factor: \( x - 3 = 0 \)
- To satisfy this factor, set \( x - 3 = 0 \).
- Solving for \( x \), we get \( x = 3 \).
2. Factor: \( x^2 + 4 = 0 \)
- To satisfy this factor, set \( x^2 + 4 = 0 \).
- Solving for \( x \), we get \( x^2 = -4 \).
- Taking the square root of both sides, we find \( x = \pm 2i \) (complex solutions).
3. Factor: \( x + 1 = 0 \)
- To satisfy this factor, set \( x + 1 = 0 \).
- Solving for \( x \), we get \( x = -1 \).
Now, let's summarize:
- The equation has three solutions: \( x = 3, x = 2i, x = -2i, \) and \( x = -1 \).
- Among these solutions, the real solutions are \( x = 3 \) and \( x = -1 \).
So, the completed statement should be:
The equation has [tex]\( \boxed{3} \)[/tex] solutions. Its real solutions are [tex]\( x = \boxed{3, -1} \)[/tex].
