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Consider this polynomial equation:

[tex]\[ 6(x-3)\left(x^2+4\right)(x+1)=0 \][/tex]

Use the equation to complete this statement:
The equation has [tex]$\square$[/tex] solutions. Its real solutions are [tex]$x=$[/tex] [tex]$\square$[/tex] .

Sagot :

GinnyAnswer
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].
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