show all the steps to write a polynomial from (x-2+sq root 7) times (x-2+sq root 7)

We are tasked with multiplying \((x - 2 + \sqrt{7}) \times (x - 2 + \sqrt{7})\) to form a polynomial. This is essentially the same as squaring the binomial \( (x - 2 + \sqrt{7})^2 \).

### Step-by-Step Solution:

1. **Express the multiplication:**
   \[
   (x - 2 + \sqrt{7}) \times (x - 2 + \sqrt{7}) = (x - 2 + \sqrt{7})^2
   \]

2. **Use the formula for squaring a trinomial:**
   \[
   (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc
   \]
   In our case, let:
   - \(a = x\),
   - \(b = -2\),
   - \(c = \sqrt{7}\).

3. **Square each term individually:**
   \[
   a^2 = x^2
   \]
   \[
   b^2 = (-2)^2 = 4
   \]
   \[
   c^2 = (\sqrt{7})^2 = 7
   \]

4. **Calculate the cross terms:**
   - \(2ab = 2 \times x \times (-2) = -4x\),
   - \(2ac = 2 \times x \times \sqrt{7} = 2x\sqrt{7}\),
   - \(2bc = 2 \times (-2) \times \sqrt{7} = -4\sqrt{7}\).

5. **Combine all the terms:**
   \[
   (x - 2 + \sqrt{7})^2 = x^2 + 4 + 7 + (-4x) + 2x\sqrt{7} + (-4\sqrt{7})
   \]

6. **Simplify the constants and terms:**
   - \(x^2\) remains,
   - Combine constants: \(4 + 7 = 11\),
   - The other terms remain as they are.

   So, the final expanded polynomial is:
   \[
   x^2 - 4x + 2x\sqrt{7} - 4\sqrt{7} + 11
   \]

### Final Answer:
The expanded polynomial is:
\[
x^2 - 4x + 2x\sqrt{7} - 4\sqrt{7} + 11
\]

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Would you like more details or further clarification?

Here are 5 related questions:
1. How do you multiply binomials with irrational numbers?
2. Can this polynomial be factored further?
3. What is the geometric significance of the square root term in this polynomial?
4. How would the process change if \(\sqrt{7}\) were replaced by another constant?
5. What are the steps to graph this polynomial function?

**Tip:** Always group similar terms together after multiplying expressions to simplify the final polynomial.

Follow this detailed guide to expand and simplify the expression (x - 2 + sqrt(7))^2. Learn how to apply the trinomial squaring formula and work with irrational numbers in this algebraic solution. Suitable for students in Grades 9-11.

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