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factor ( x^4 - 2x^2 + 1 ) completely. all factors in your answer should…

Question

factor ( x^4 - 2x^2 + 1 ) completely. all factors in your answer should have integer coefficients.

Explanation:

Step1: Recognize the quadratic in \(x^2\)

The expression \(x^4 - 2x^2 + 1\) can be seen as a quadratic in terms of \(y = x^2\). So we rewrite it as \(y^2 - 2y + 1\), where \(y=x^2\).

Step2: Factor the quadratic

Using the perfect square trinomial formula \(a^2 - 2ab + b^2=(a - b)^2\), for \(y^2 - 2y + 1\), we have \(a = y\) and \(b = 1\). So \(y^2 - 2y + 1=(y - 1)^2\).

Step3: Substitute back \(y = x^2\)

Substituting \(y = x^2\) back into \((y - 1)^2\), we get \((x^2 - 1)^2\).

Step4: Factor \(x^2 - 1\) further

Using the difference of squares formula \(a^2 - b^2=(a - b)(a + b)\), for \(x^2 - 1\), we have \(a = x\) and \(b = 1\). So \(x^2 - 1=(x - 1)(x + 1)\).

Step5: Write the final factored form

Substituting \(x^2 - 1=(x - 1)(x + 1)\) into \((x^2 - 1)^2\), we get \([(x - 1)(x + 1)]^2=(x - 1)^2(x + 1)^2\).

Answer:

\((x - 1)^2(x + 1)^2\)