Factor completely. \[ x^{3}+6 x^{2}-15 x-90 \]

To factor the polynomial \( x^{3}+6 x^{2}-15 x-90 \) completely, we first look for possible rational roots using the Rational Root Theorem. Testing \( x = -3 \): \[ (-3)^3 + 6(-3)^2 - 15(-3) - 90 = -27 + 54 + 45 - 90 = -18 \quad (\text{Not a root}) \] Next, let's try \( x = -5 \): \[ (-5)^3 + 6(-5)^2 - 15(-5) - 90 = -125 + 150 + 75 - 90 = 10 \quad (\text{Not a root}) \] Now, testing \( x = -6 \): \[ (-6)^3 + 6(-6)^2 - 15(-6) - 90 = -216 + 216 + 90 - 90 = 0 \quad (\text{It is a root!}) \] Now that we know \( x = -6 \) is a root, we can divide the polynomial by \( x + 6 \) using synthetic division: \[ \begin{array}{r|rrrr} -6 & 1 & 6 & -15 & -90 \\ & & -6 & 0 & 90 \\ \hline & 1 & 0 & -15 & 0 \\ \end{array} \] This gives us the quotient \( x^2 - 15 \). Now, we factor \( x^2 - 15 \) further: \[ x^2 - 15 = (x - \sqrt{15})(x + \sqrt{15}) \] Putting everything together, we have: \[ x^{3}+6 x^{2}-15 x-90 = (x + 6)(x - \sqrt{15})(x + \sqrt{15}) \] So, the completely factored form is: \[ (x + 6)(x - \sqrt{15})(x + \sqrt{15}) \]