Find all the zeros. Write the answer in exact form. \( p(x)=x^{4}-10 x^{2}-56 \) If there is more than one answer, separate them with commas. Select "None" if applicable. The zeros of \( p(x): \square \)

To find the zeros of the polynomial \( p(x) = x^{4} - 10x^{2} - 56 \), we can start by making a substitution. Let \( y = x^{2} \). This turns the equation into a quadratic form: \( p(y) = y^{2} - 10y - 56 \). Now we can solve for \( y \) using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{10 \pm \sqrt{(-10)^{2} - 4(1)(-56)}}{2(1)} = \frac{10 \pm \sqrt{100 + 224}}{2} = \frac{10 \pm \sqrt{324}}{2} = \frac{10 \pm 18}{2}. \] This gives us two values for \( y \): 1. \( y = \frac{28}{2} = 14 \) 2. \( y = \frac{-8}{2} = -4 \) (Not valid since \( y = x^2 \)) Now we back substitute \( y = x^2 \): 1. For \( y = 14 \): \[ x^2 = 14 \implies x = \pm \sqrt{14}. \] Since \( y = -4 \) is not valid, we only have \( x = \sqrt{14}, -\sqrt{14} \). Thus, the zeros of \( p(x) \) are: \[ \sqrt{14}, -\sqrt{14}. \]