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 $$ p(x): \square $$

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We start with the polynomial
$$
p(x)=x^4-10x^2-56.
$$
Notice that the polynomial is quadratic in form if we let $$ y=x^2 $$. Substituting, we have
$$
y^2-10y-56=0.
$$

To solve for $$ y $$, we use the quadratic formula:
$$
y=\frac{-b \pm \sqrt{b^2-4ac}}{2a},
$$
where $$ a=1 $$, $$ b=-10 $$, and $$ c=-56 $$. Thus,
$$
y=\frac{-(-10) \pm \sqrt{(-10)^2-4(1)(-56)}}{2(1)}=\frac{10\pm \sqrt{100+224}}{2}=\frac{10\pm \sqrt{324}}{2}.
$$
Since $$\sqrt{324}=18$$, we obtain
$$
y=\frac{10\pm 18}{2}.
$$

This gives us two solutions for $$ y $$:
$$
y=\frac{10+18}{2}=14 \quad \text{and} \quad y=\frac{10-18}{2}=-4.
$$

Recall that $$ y=x^2 $$. We now solve for $$ x $$ from each equation.

1. For $$ x^2=14 $$,
   $$
   x=\sqrt{14} \quad \text{or} \quad x=-\sqrt{14}.
   $$

2. For $$ x^2=-4 $$,
   $$
   x=\sqrt{-4}=2i \quad \text{or} \quad x=-2i.
   $$

Thus, the zeros of $$ p(x) $$ are:
$$
\sqrt{14},\ -\sqrt{14},\ 2i,\ -2i.
$$
The zeros of $$ p(x) $$ are $$ \sqrt{14}, -\sqrt{14}, 2i, -2i $$.

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 \( p(x): \square \)

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