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Multiplying Binomials with Radicals

When you multiply a square root by itself, the square root disappears.

$\sqrt{7} \times \sqrt{7}=7$

This leads to a pattern when working with $i$ (which equals $\sqrt{-1}$):

$i^1=\sqrt{-1}$

$i^2=\sqrt{-1} \times \sqrt{-1} =-1$

$i^3=\sqrt{-1} \times \sqrt{-1} \times \sqrt{-1} =-\sqrt{-1}  $

$i^4=\sqrt{-1} \times \sqrt{-1} \times \sqrt{-1} \times \sqrt{-1}= -1 \times -1 = 1$

You have already learned this, but it starts to get fun when you put radicals (and $i$) in binomials and multiply (remember these FOIL problems?).  

$\eqalign{(x+\sqrt{2})(x+\sqrt{2})\\x^2+2x\sqrt{2}+2}$

See how the last term from that FOIL problem lost its square root sign?

That becomes even more fun when radicals are used in special products.  What do you think will happen here?

$\eqalign{(x+\sqrt{2})(x-\sqrt{2})}$

That's right!  The middle terms cancel each other out, so the answer is $x^2-2$.

Standardized tests love these types of multiplication problems because they look complicated but they are fairly simple -- and the weird symbols (radicals, $i$s) tend to fall out of the problems. 

Now, the radicals don't always fall out.  Sometimes, they don't get cancelled out.  So pay attention.  Also, remember the basic rules of multiplying radicals: if the radicals don't cancel each other out, you can multiply numbers under radicals:

$\sqrt{8} \times \sqrt{10}= \sqrt{80}$

This works when radicals are simplified as well:

$2\sqrt{2} \times \sqrt{10}= 2\sqrt{20} = 2 \times 2 \sqrt{5}=4\sqrt{5}$

(Note: these two examples are the same, the first is not simplified and the second is.  You can always multiply the numbers under radicals and keep the answer under the radical. If you have coefficients on the radicals (like the 4 on the $\sqrt{5}$ in $4\sqrt{5}$), you can multiply the coefficients, just keep them outside the radicals. 

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