Question

1. $x^{4}-12x^{2}+36$

Explanation:

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

Treat \(x^4 - 12x^2 + 36\) as a quadratic: let \(y = x^2\), so it becomes \(y^2 - 12y + 36\).

Step2: Factor the quadratic

Factor \(y^2 - 12y + 36\) using \((a - b)^2 = a^2 - 2ab + b^2\). Here, \(a = y\), \(2ab = 12y\) (so \(b = 6\)), so \(y^2 - 12y + 36=(y - 6)^2\).

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

We get \((x^2 - 6)^2\).

Step4: Factor \(x^2 - 6\) (optional, if difference of squares)

\(x^2 - 6=(x - \sqrt{6})(x + \sqrt{6})\), so overall \((x - \sqrt{6})^2(x + \sqrt{6})^2\). But the initial factoring to \((x^2 - 6)^2\) is also a valid factored form.

Answer:

\(\boxed{(x^2 - 6)^2}\) (or \(\boxed{(x - \sqrt{6})^2(x + \sqrt{6})^2}\))