Certainly! To expand the expression \((1+\sqrt{2})(3-\sqrt{2})\) and give the answer in the form \(a + b\sqrt{2}\), we will proceed step-by-step using the distributive property.
First, let's distribute each term in the first binomial \((1+\sqrt{2})\) by each term in the second binomial \((3-\sqrt{2})\):
[tex]\[
(1 + \sqrt{2})(3 - \sqrt{2}) = 1 \cdot 3 + 1 \cdot (-\sqrt{2}) + \sqrt{2} \cdot 3 + \sqrt{2} \cdot (-\sqrt{2})
\][/tex]
Now, let's evaluate each term individually:
1. Multiply the constants:
[tex]\[
1 \cdot 3 = 3
\][/tex]
2. Multiply the constant by the negative square root term:
[tex]\[
1 \cdot -\sqrt{2} = -\sqrt{2}
\][/tex]
3. Multiply the square root term by the constant:
[tex]\[
\sqrt{2} \cdot 3 = 3\sqrt{2}
\][/tex]
4. Multiply the square root terms:
[tex]\[
\sqrt{2} \cdot (-\sqrt{2}) = -(\sqrt{2})^2 = -2
\][/tex]
Next, combine all these terms together:
[tex]\[
3 + (-\sqrt{2}) + 3\sqrt{2} + (-2)
\][/tex]
Now, let's group the integer terms and the square root terms:
1. Combine the integer terms:
[tex]\[
3 + (-2) = 1
\][/tex]
2. Combine the square root terms:
[tex]\[
-\sqrt{2} + 3\sqrt{2} = 2\sqrt{2}
\][/tex]
So, the expanded expression, combining both integer and square root parts, is:
[tex]\[
1 + 2\sqrt{2}
\][/tex]
Thus, the expression \((1+\sqrt{2})(3-\sqrt{2})\) in the form \(a + b\sqrt{2}\) is:
[tex]\[
\boxed{1 + 2\sqrt{2}}
\][/tex]
