Question

question
factor completely over the integers.
$x^4 - 6x^2 + 5$

Explanation:

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

Substitute \( y = x^2 \) into \( x^4 - 6x^2 + 5 \), we get \( y^2 - 6y + 5 \).

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

We need two numbers that multiply to \( 5 \) and add to \( -6 \). The numbers are \( -1 \) and \( -5 \). So, \( y^2 - 6y + 5=(y - 1)(y - 5) \).

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

We have \( (x^2 - 1)(x^2 - 5) \).

Step4: Factor \( x^2 - 1 \) (difference of squares).

Using the formula \( a^2 - b^2=(a - b)(a + b) \), \( x^2 - 1=(x - 1)(x + 1) \). The \( x^2 - 5 \) doesn't factor over the integers. So the completely factored form is \( (x - 1)(x + 1)(x^2 - 5) \).

Answer:

\((x - 1)(x + 1)(x^2 - 5)\)