# If (3-sqrt2) and (2+sqrt2) are two of the roots of a fourth-degree polynomial with integer coefficients, what is the product of the other two roots?

Dec 25, 2016

$4 - \sqrt{2}$

#### Explanation:

The two other roots will be the radical conjugates of the given ones, namely:

$\left(3 + \sqrt{2}\right)$ and $\left(2 - \sqrt{2}\right)$

So (using FOIL) their product is:

$\left(3 + \sqrt{2}\right) \left(2 - \sqrt{2}\right) = {\overbrace{3 \cdot 2}}^{\text{First"+overbrace(3(-sqrt(2)))^"Outside"+overbrace((sqrt(2))2)^"Inside"+overbrace((sqrt(2))(-sqrt(2)))^"Last}}$

$\textcolor{w h i t e}{\left(3 + \sqrt{2}\right) \left(2 - \sqrt{2}\right)} = 6 - 3 \sqrt{2} + 2 \sqrt{2} - 2$

$\textcolor{w h i t e}{\left(3 + \sqrt{2}\right) \left(2 - \sqrt{2}\right)} = 4 - \sqrt{2}$