Question

question
factor completely over the integers.
$x^4 - 12x^2 + 32$

Explanation:

Step1: Let \( y = x^2 \), rewrite the polynomial.

We substitute \( y = x^2 \) into the given polynomial \( x^4 - 12x^2 + 32 \), so it becomes \( y^2 - 12y + 32 \).

Step2: Factor the quadratic in \( y \).

We need to find two numbers that multiply to \( 32 \) and add up to \( - 12 \). The numbers are \( - 4 \) and \( - 8 \). So, \( y^2 - 12y + 32=(y - 4)(y - 8) \).

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

Substituting \( y = x^2 \) back, we get \( (x^2 - 4)(x^2 - 8) \).

Step4: Factor \( x^2 - 4 \) further.

Using the difference of squares formula \( a^2 - b^2=(a + b)(a - b) \), where \( a = x \) and \( b = 2 \), we have \( x^2 - 4=(x + 2)(x - 2) \). The term \( x^2 - 8 \) cannot be factored over the integers.

Answer:

\((x + 2)(x - 2)(x^2 - 8)\)