To factor the polynomial \( 2x^5 - 4x^3 + 2x \), we can start by factoring out the greatest common factor from all the terms.
The greatest common factor here is \( 2x \). So we factor \( 2x \) out:
\[ 2x^5 - 4x^3 + 2x = 2x(x^4 - 2x^2 + 1) \]
Now, we need to factor the remaining polynomial \( x^4 - 2x^2 + 1 \). We can recognize that \( x^4 - 2x^2 + 1 \) is a quadratic in terms of \( x^2 \). Let's make a substitution where \( u = x^2 \):
\[ u^2 - 2u + 1 \]
This can be factored as:
\[ (u - 1)^2 = (x^2 - 1)^2 \]
Now substituting back \( u = x^2 \):
\[ x^4 - 2x^2 + 1 = (x^2 - 1)^2 \]
The expression \( x^2 - 1 \) can be factored further using the difference of squares:
\[ x^2 - 1 = (x - 1)(x + 1) \]
Thus, we have:
\[ (x^2 - 1)^2 = ((x - 1)(x + 1))^2 = (x - 1)^2(x + 1)^2 \]
Putting everything together, we can rewrite the original polynomial as:
\[ 2x^5 - 4x^3 + 2x = 2x((x - 1)^2(x + 1)^2) \]
Setting this equal to zero to solve the polynomial equation \( 2x^5 - 4x^3 + 2x = 0 \):
\[ 2x((x - 1)^2(x + 1)^2) = 0 \]
This equation is satisfied if any of the factors equal zero:
- \( 2x = 0 \) implies \( x = 0 \)
- \( (x - 1)^2 = 0 \) implies \( x = 1 \) (with a multiplicity of 2)
- \( (x + 1)^2 = 0 \) implies \( x = -1 \) (with a multiplicity of 2)
Thus, the solutions to the equation \( 2x^5 - 4x^3 + 2x = 0 \) are:
\[ x = 0, \quad x = 1 \quad \text{(multiplicity 2)}, \quad x = -1 \quad \text{(multiplicity 2)} \]