Suppose that $$ \sqrt{5} $$ is a zero of $$ p(x)=x^{3}-\sqrt{5} x^{2}-25 x+25 \sqrt{5} $$ and $$ q(x)=x^{3}-5 x^{2}-5 x+25 $$

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We start with the two polynomials

$$
p(x)=x^3-\sqrt{5}\,x^2-25x+25\sqrt{5} \quad \text{and} \quad q(x)=x^3-5x^2-5x+25.
$$

It is given that $$\sqrt{5}$$ is a zero of both $$p(x)$$ and $$q(x)$$. We will factor each polynomial step by step.

---

### Factoring $$ p(x) $$

Since $$\sqrt{5}$$ is a zero, the linear factor $$(x-\sqrt{5})$$ divides $$p(x)$$. Write

$$
p(x)=(x-\sqrt{5})(Ax^2+Bx+C).
$$

Expand the right-hand side:

$$
(x-\sqrt{5})(Ax^2+Bx+C) = Ax^3+Bx^2+Cx - A\sqrt{5}\,x^2 -B\sqrt{5}\,x - C\sqrt{5}.
$$

Collect like terms:

$$
= Ax^3+\left(B-A\sqrt{5}\right)x^2+\left(C-B\sqrt{5}\right)x - C\sqrt{5}.
$$

This must equal

$$
x^3-\sqrt{5}\,x^2-25x+25\sqrt{5}.
$$

By equating coefficients:

1. Coefficient of $$x^3$$: 
   $$
   A=1.
   $$

2. Coefficient of $$x^2$$:
   $$
   B-A\sqrt{5} = B-\sqrt{5} = -\sqrt{5} \quad\Longrightarrow\quad B=0.
   $$

3. Coefficient of $$x$$:
   $$
   C-B\sqrt{5} = C-0 = -25 \quad\Longrightarrow\quad C=-25.
   $$

4. Constant term:
   $$
   -C\sqrt{5} = -(-25)\sqrt{5} = 25\sqrt{5},
   $$
   which is consistent.

Thus, the factorization is

$$
p(x) = (x-\sqrt{5})(x^2-25).
$$

Notice that the quadratic factors further:

$$
x^2-25 = (x-5)(x+5).
$$

So the complete factorization is

$$
p(x) = (x-\sqrt{5})(x-5)(x+5).
$$

The zeros of $$p(x)$$ are

$$
x=\sqrt{5},\quad x=5,\quad x=-5.
$$

---

### Factoring $$ q(x) $$

Similarly, since $$\sqrt{5}$$ is a zero of $$q(x)$$, we have that $$(x-\sqrt{5})$$ is a factor. We start with

$$
q(x)=(x-\sqrt{5})(Ax^2+Bx+C).
$$

Expanding gives:

$$
(x-\sqrt{5})(Ax^2+Bx+C) = Ax^3+\left(B-A\sqrt{5}\right)x^2+\left(C-B\sqrt{5}\right)x - C\sqrt{5}.
$$

This must equal

$$
x^3-5x^2-5x+25.
$$

Matching coefficients:

1. Coefficient of $$x^3$$:
   $$
   A=1.
   $$

2. Coefficient of $$x^2$$:
   $$
   B-A\sqrt{5} = B-\sqrt{5} = -5 \quad\Longrightarrow\quad B = -5+\sqrt{5}.
   $$

3. Coefficient of $$x$$:
   $$
   C - B\sqrt{5} = -5.
   $$
   Substitute $$B=-5+\sqrt{5}$$:
   $$
   C - \sqrt{5}(-5+\sqrt{5}) = C + 5\sqrt{5} - 5 = -5 \quad\Longrightarrow\quad C = -5 - 5\sqrt{5} + 5 = -5\sqrt{5}.
   $$

4. Constant term:
   $$
   -C\sqrt{5} = -(-5\sqrt{5})\sqrt{5} = 5\sqrt{5}\cdot \sqrt{5} =5\cdot5=25,
   $$
   which is correct.

Thus, we have

$$
q(x) = (x-\sqrt{5})\Bigl(x^2+\bigl(-5+\sqrt{5}\bigr)x-5\sqrt{5}\Bigr).
$$

However, we notice something interesting: we test if $$-\sqrt{5}$$ is a zero for $$q(x)$$. Substitute $$x=-\sqrt{5}$$ into $$q(x)$$:

$$
q(-\sqrt{5}) = (-\sqrt{5})^3-5(-\sqrt{5})^2-5(-\sqrt{5})+25.
$$

Calculate each term:

$$
(-\sqrt{5})^3 = -5\sqrt{5},\quad (-\sqrt{5})^2 =5,\quad -5(-\sqrt{5}) = 5\sqrt{5}.
$$

Thus,

$$
q(-\sqrt{5}) = -5\sqrt{5}-5(5)+5\sqrt{5}+25 = -5\sqrt{5}-25+5\sqrt{5}+25 = 0.
$$

So $$(x+\sqrt{5})$$ is also a factor of $$q(x)$$.

Since we already have the factor $$(x-\sqrt{5})$$, we combine $$(x-\sqrt{5})(x+\sqrt{5})$$ to get

$$
x^2-5.
$$

Now we factor $$q(x)$$ by dividing by $$x^2-5$$:

Divide

$$
q(x)=x^3-5x^2-5x+25
$$
by
$$
x^2-5.
$$

Perform the division:

1. $$x^3 \div x^2 = x$$. Multiply: $$x(x^2-5) = x^3-5x$$.
2. Subtract: $$(x^3-5x^2-5x+25) - (x^3-5x) = -5x^2+0x+25$$.
3. Now, $$-5x^2 \div x^2 = -5$$. Multiply: $$-5(x^2-5) = -5x^2+25$$.
4. Subtract: $$(-5x^2+0x+25) - (-5x^2+25) =0$$.

So,

$$
q(x)=(x^2-5)(x-5).
$$

And since

$$
x^2-5 = (x-\sqrt{5})(x+\sqrt{5}),
$$

the complete factorization is

$$
q(x)=(x-\sqrt{5})(x+\sqrt{5})(x-5).
$$

The zeros of $$q(x)$$ are

$$
x=\sqrt{5},\quad x=-\sqrt{5},\quad x=5.
$$

---

### Final Answers

$$
p(x) = (x-\sqrt{5})(x-5)(x+5) \quad \text{with zeros } x=\sqrt{5},\,5,\,-5,
$$

$$
q(x) = (x-\sqrt{5})(x+\sqrt{5})(x-5) \quad \text{with zeros } x=\sqrt{5},\,-\sqrt{5},\,5.
$$
The polynomials factor as follows: - $$ p(x) = (x - \sqrt{5})(x - 5)(x + 5) $$ with zeros at $$ x = \sqrt{5}, 5, -5 $$. - $$ q(x) = (x - \sqrt{5})(x + \sqrt{5})(x - 5) $$ with zeros at $$ x = \sqrt{5}, -\sqrt{5}, 5 $$.

Suppose that is a zero of and

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