To solve the polynomial equation \( x^4 - 3x^2 + 2 = 0 \) using factoring, we can first make a substitution to simplify the expression. Let's set \( y = x^2 \). This gives us:
\[ y^2 - 3y + 2 = 0 \]
Next, we can factor this quadratic equation. We look for two numbers that multiply to \( 2 \) (the constant term) and add to \( -3 \) (the coefficient of the \( y \) term). The numbers \( -1 \) and \( -2 \) fit these criteria:
\[ (y - 1)(y - 2) = 0 \]
Now, we can solve for \( y \):
- \( y - 1 = 0 \) leads to \( y = 1 \)
- \( y - 2 = 0 \) leads to \( y = 2 \)
Recalling that \( y = x^2 \), we substitute back to find values for \( x \):
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For \( y = 1 \): \[ x^2 = 1 \implies x = \pm 1 \]
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For \( y = 2 \): \[ x^2 = 2 \implies x = \pm \sqrt{2} \]
Putting it all together, the solutions for the original polynomial equation \( x^4 - 3x^2 + 2 = 0 \) are:
\[ x = 1, , x = -1, , x = \sqrt{2}, , x = -\sqrt{2} \]
So the final answer is:
\[ x = 1, -1, \sqrt{2}, -\sqrt{2} \]