# Find the product of the pair of conjugates. (2sqrt(3) -sqrt(7))(2sqrt(3)+sqrt(7))

Find the product of the pair of conjugates.
$\left(2\sqrt{3}-\sqrt{7}\right)\left(2\sqrt{3}+\sqrt{7}\right)$
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For this question, we are going to apply the same method we use inpolynomials called the FOIL Method which states:
First
Outer
Inner
Last
Therefore, let's follow this process to get the following:
$\left[2\sqrt{3}\right]\left[2\sqrt{3}\right]+\left[2\sqrt{3}\right]\left[\sqrt{7}\right]+\left[-2\sqrt{3}\right]\left[\sqrt{7}\right]-\left[\sqrt{7}\right]\left[\sqrt{7}\right]$
Now, whenever we're multiplying the same radical as seen above, wecan simplify it as just the number inside the radical beingmultiplied by the constants to get the following:
$4\left(3\right)+2\sqrt{3}\sqrt{7}-2\sqrt{3}\sqrt{7}-7$
With further simplification from the above we can get the followingas a result:
FINAL ANSWER: 12 - 7 = 5
###### Not exactly what you’re looking for?
$\left(2\sqrt{3}-\sqrt{7}\right)\left(2\sqrt{3}+\sqrt{7}\right)$, FOIL method, this look like$\left(a-b\right)\left(a-b\right)={a}^{2}-{b}^{2}$
$=\left(2\sqrt{3}\right)\left(2\sqrt{3}\right)-2\sqrt{3}\ast \sqrt{7}+2\sqrt{3}\ast \sqrt{7}-\sqrt{7}\ast \sqrt{7}$
=4(3)-7
=5