We are asked to factor the expression $7x^4 - 13x^2 - 2$ completely.

1. Problem Description

We are asked to factor the expression $7x^4 - 13x^2 - 2$ completely.

2. Solution Steps

Let $y = x^2$. Then the expression becomes $7y^2 - 13y - 2$. We are looking for two numbers that multiply to $7(-2) = -14$ and add to $-13$. Those two numbers are $-14$ and $1$. We rewrite the middle term as $-14y + y$, so the expression becomes $7y^2 - 14y + y - 2$. We factor by grouping:

$7y(y - 2) + 1(y - 2) = (7y + 1)(y - 2)$.

Now, we substitute $x^2$ back in for $y$:

$(7x^2 + 1)(x^2 - 2)$.

The expression $7x^2 + 1$ cannot be factored further using real numbers because $7x^2 + 1 = 0$ means $x^2 = -\frac{1}{7}$, so $x = \pm \sqrt{-\frac{1}{7}}$, which are complex numbers. The expression $x^2 - 2$ can be factored as $(x - \sqrt{2})(x + \sqrt{2})$.

So $7x^4 - 13x^2 - 2 = (7x^2 + 1)(x^2 - 2) = (7x^2 + 1)(x-\sqrt{2})(x+\sqrt{2})$. The problem probably intended for integer factoring, so the answer is $(7x^2 + 1)(x^2 - 2)$.

3. Final Answer

$(7x^2+1)(x^2-2)$