Question

question 27
identify each factor of
$x^4 - 16$
$\square\\ x + 2$
$\square\\ x - 2$
$\square\\ x^2 + 4$
$\square\\ x^2 - 4$

Explanation:

Step1: Factor \(x^4 - 16\) using difference of squares

Recall that \(a^2 - b^2=(a + b)(a - b)\). Here, \(x^4 - 16=(x^2)^2-4^2\), so we can factor it as \((x^2 + 4)(x^2 - 4)\).

Step2: Factor \(x^2 - 4\) further

Again, using the difference of squares formula, \(x^2 - 4=(x + 2)(x - 2)\).
So the factors of \(x^4 - 16\) are \(x + 2\), \(x - 2\), \(x^2 + 4\), and \(x^2 - 4\) (wait, in the options, we have \(x^2+4\) and the linear factors \(x + 2\), \(x - 2\), and let's check the options: the options are \(x + 2\), \(x - 2\), \(x^2+4\), and another (maybe a typo, but from the factoring, \(x^4-16=(x^2 + 4)(x + 2)(x - 2)\), so the factors among the options are \(x + 2\), \(x - 2\), \(x^2+4\).

Answer:

A. \(x + 2\)
B. \(x - 2\)
C. \(x^2 + 4\)