To apply the Fundamental Theorem of Algebra to the equation, we first need to simplify and rearrange it into a standard polynomial form. The given equation is:
\[ 12x - 6x^2 + 3x^4 = 6x^3 + 2x - x^4 \]
First, let's move all terms to one side of the equation:
\[ 3x^4 + x^4 - 6x^3 - 6x^2 + 12x - 2x = 0 \]
Combining like terms gives:
\[ (3x^4 + x^4) - 6x^3 - 6x^2 + (12x - 2x) = 0 \]
\[ 4x^4 - 6x^3 - 6x^2 + 10x = 0 \]
Now, we can factor out the greatest common factor from the left-hand side:
\[ 2x(2x^3 - 3x^2 - 3x + 5) = 0 \]
Setting \(2x = 0\) gives us one root:
\[ x = 0 \]
Now we need to find the roots of the cubic polynomial \(2x^3 - 3x^2 - 3x + 5 = 0\).
The Fundamental Theorem of Algebra states that a polynomial of degree \(n\) has exactly \(n\) roots (counting multiplicities) in the complex numbers. The cubic polynomial has a degree of 3, which means it will have 3 roots in total.
Combining both parts, we have:
- One root from \(2x = 0\): \(x = 0\)
- Three roots from the cubic polynomial \(2x^3 - 3x^2 - 3x + 5 = 0\)
Thus, the total number of roots for the equation \(12x - 6x^2 + 3x^4 = 6x^3 + 2x - x^4\) is:
\[ 1 + 3 = 4 \]
Therefore, the equation has 4 roots in total, counting multiplicities.
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