Factor \( -6 x^{4}+8 x^{2}+8 \) completely

To factor the expression \(-6 x^{4}+8 x^{2}+8\), we start by looking for a common factor in all the terms. We can factor out \(-2\): \[ -2(3x^4 - 4x^2 - 4) \] Next, we notice that the expression inside the parentheses is a quadratic in terms of \(x^2\). Let's substitute \(y = x^2\), then we have: \[ -2(3y^2 - 4y - 4) \] Next, we can factor \(3y^2 - 4y - 4\). To do this, we look for two numbers that multiply to \(3 \times -4 = -12\) and add to \(-4\). Those numbers are \(-6\) and \(2\). We can rewrite the middle term: \[ 3y^2 - 6y + 2y - 4 \] Now factor by grouping: \[ = 3y(y - 2) + 2(y - 2) \] This gives: \[ = (3y + 2)(y - 2) \] Substituting back \(y = x^2\), we get: \[ = (3x^2 + 2)(x^2 - 2) \] Putting it all together, we have: \[ = -2(3x^2 + 2)(x^2 - 2) \] Next, we can factor \(x^2 - 2\) further into \((x - \sqrt{2})(x + \sqrt{2})\): Therefore, the completely factored form of \(-6 x^{4}+8 x^{2}+8\) is: \[ -2(3x^2 + 2)(x - \sqrt{2})(x + \sqrt{2}) \]