To factor the polynomial equation \(2x^5 - 4x^3 + 2x = 0\), we can start by factoring out the common term.
First, observe that each term in the polynomial has a common factor:
\[ 2x^5 - 4x^3 + 2x = 2x(x^4 - 2x^2 + 1) \]
Now, we are left with the factorization:
\[ 2x(x^4 - 2x^2 + 1) = 0 \]
Next, we need to solve for \(x\) by setting each factor equal to zero:
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From \(2x = 0\): \[ x = 0 \]
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Now, we solve \(x^4 - 2x^2 + 1 = 0\). Let's make a substitution to simplify this. Let \(y = x^2\). Then the equation becomes:
\[ y^2 - 2y + 1 = 0 \]
This factors as:
\[ (y - 1)^2 = 0 \]
Hence, we have:
\[ y - 1 = 0 \implies y = 1 \]
Substituting back \(y = x^2\), we get:
\[ x^2 = 1 \]
This gives us two solutions:
\[ x = 1 \quad \text{or} \quad x = -1 \]
Now, we can summarize the solutions for the original equation:
\[ x = 0, \quad x = 1, \quad x = -1 \]
Thus, the complete factored form of the polynomial is:
\[ 2x(x - 1)^2(x + 1)^2 = 0 \]
The solutions are:
\[ x = 0, \quad x = 1, \quad x = -1 \]