Question

question
factor completely over the integers.
$x^{4}-3x^{2}-18$

Explanation:

Step1: Substitute $u=x^2$

Let $u = x^2$, so the polynomial becomes:
$u^2 - 3u - 18$

Step2: Factor the quadratic in $u$

Find two integers that multiply to $-18$ and add to $-3$: $-6$ and $3$.
$u^2 - 3u - 18 = (u - 6)(u + 3)$

Step3: Substitute back $u=x^2$

Replace $u$ with $x^2$:
$(x^2 - 6)(x^2 + 3)$

Step4: Check for further integer factorization

$x^2 - 6$ and $x^2 + 3$ cannot be factored into integer linear terms, so this is the complete factorization over integers.

Answer:

$(x^2 - 6)(x^2 + 3)$