Question

Apply the Fundamental Theorem of Algebra to find the number of roots for the equation 12x−6x2+3x4=6x3+2x−x4.(1 point)

Answer 1

To apply the Fundamental Theorem of Algebra, we first need to simplify the given equation:

\[
12x - 6x^2 + 3x^4 = 6x^3 + 2x - x^4.
\]

Next, let's move all terms to one side of the equation:

\[
3x^4 + x^4 - 6x^3 - 6x^2 + 12x - 2x = 0.
\]

Combining the like terms gives us:

\[
(3x^4 + x^4) - 6x^3 - 6x^2 + (12x - 2x) = 0,
\]
\[
4x^4 - 6x^3 - 6x^2 + 10x = 0.
\]

Now we can factor out the greatest common factor, which is 2x:

\[
2x(2x^3 - 3x^2 - 3x + 5) = 0.
\]

Now we can see that \(x = 0\) is one root. Next, we need to determine the number of roots of the cubic polynomial \(2x^3 - 3x^2 - 3x + 5\).

According to the Fundamental Theorem of Algebra, a polynomial of degree \(n\) has exactly \(n\) roots in the complex number system (counting multiplicities). The cubic polynomial \(2x^3 - 3x^2 - 3x + 5\) has a degree of 3.

Therefore, it will have a total of 3 roots (counting both real and complex roots).

Combining this with the root we factored out, the total number of roots for the original equation \(12x - 6x^2 + 3x^4 - (6x^3 + 2x - x^4) = 0\) is:

1 (from \(2x\)) + 3 (from \(2x^3 - 3x^2 - 3x + 5\)) = 4 roots.

Thus, the total number of roots for the original equation is **4**.