This is factored into (1 + sqrt(2))(1 + sqrt(3))
explain how

Let's explore how the expression \(\sqrt{6} + \sqrt{3} + \sqrt{2} + 1\) can be factored into \((1 + \sqrt{2})(1 + \sqrt{3})\).

### Step 1: Expand the product
Start by expanding the expression \((1 + \sqrt{2})(1 + \sqrt{3})\):

\[
(1 + \sqrt{2})(1 + \sqrt{3}) = 1 \cdot 1 + 1 \cdot \sqrt{3} + \sqrt{2} \cdot 1 + \sqrt{2} \cdot \sqrt{3}
\]

### Step 2: Simplify the expansion
Simplify each term:

\[
1 \cdot 1 = 1
\]
\[
1 \cdot \sqrt{3} = \sqrt{3}
\]
\[
\sqrt{2} \cdot 1 = \sqrt{2}
\]
\[
\sqrt{2} \cdot \sqrt{3} = \sqrt{6}
\]

Now, combine all the terms:

\[
1 + \sqrt{3} + \sqrt{2} + \sqrt{6}
\]

### Step 3: Compare with the original expression
Notice that this expression is exactly the same as the original one: \(\sqrt{6} + \sqrt{3} + \sqrt{2} + 1\).

### Conclusion
This confirms that the expression \(\sqrt{6} + \sqrt{3} + \sqrt{2} + 1\) can indeed be factored into \((1 + \sqrt{2})(1 + \sqrt{3})\).

If you have any further questions or need additional details, feel free to ask!

Here are some related questions you might find interesting:

1. How do you factor expressions with multiple square roots?
2. What techniques are used to recognize when an expression can be factored?
3. Can you generalize the process for factoring other similar expressions?
4. How do you expand and simplify products of binomials?
5. What is the significance of recognizing such factorizations in higher mathematics?

**Tip:** When factoring expressions with square roots, look for patterns or identities, such as products of binomials, that might simplify the original expression.

Learn how to factorize the expression (1 + sqrt(2))(1 + sqrt(3)) into sqrt(6) + sqrt(3) + sqrt(2) + 1. This problem explores square roots and binomial expansion techniques, essential in high school mathematics.

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