To factor the polynomial \( 2x^5 - 4x^3 + 2x \), we can start by factoring out the common factor from each term.
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Identify the common factor, which is \( 2x \): \[ 2x^5 - 4x^3 + 2x = 2x(x^4 - 2x^2 + 1) \]
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Now, we need to factor the polynomial \( x^4 - 2x^2 + 1 \). We can use substitution to make this easier. Let \( y = x^2 \). Then, we can rewrite the polynomial as: \[ y^2 - 2y + 1 \]
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Now, factor \( y^2 - 2y + 1 \): \[ y^2 - 2y + 1 = (y - 1)^2 \]
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Substitute back \( y = x^2 \): \[ (x^2 - 1)^2 \]
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Recognizing that \( x^2 - 1 \) can be factored further: \[ x^2 - 1 = (x - 1)(x + 1) \]
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Thus, we have: \[ (x^2 - 1)^2 = [(x - 1)(x + 1)]^2 = (x - 1)^2(x + 1)^2 \]
Now, putting everything together, we can write: \[ 2x^5 - 4x^3 + 2x = 2x \cdot (x^2 - 1)^2 = 2x \cdot (x - 1)^2 \cdot (x + 1)^2 \]
Now we can set the equation \( 2x^5 - 4x^3 + 2x = 0 \):
\[ 2x \cdot (x - 1)^2 \cdot (x + 1)^2 = 0 \]
Thus, the solutions to the equation are:
- From \( 2x = 0 \), we get \( x = 0 \).
- From \( (x - 1)^2 = 0 \), we get \( x = 1 \) (with multiplicity 2).
- From \( (x + 1)^2 = 0 \), we get \( x = -1 \) (with multiplicity 2).
Therefore, the complete factorization is: \[ 2x(x - 1)^2(x + 1)^2 = 0 \] This gives the solutions \( x = 0, 1, -1 \).
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