How do you multiply $[\sqrt{2} + \sqrt{7}] \cdot [\sqrt{3} - \sqrt{7}]$ $[sqrt(2) + sqrt(7)]*[sqrt(3) - sqrt(7)]$ ?

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Answer 1 · 2015-07-06T21:26:12

Use the FOIL method for multiplying two binomials.

The FOIL method for multiplying two binomials indicates the order in which the terms should by multiplied.
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$(\sqrt{2} + \sqrt{7}) (\sqrt{3} - \sqrt{7})$ $(sqrt2+sqrt7)(sqrt3-sqrt7)$ =

$\sqrt{2} \sqrt{3} - \sqrt{2} \sqrt{7} + \sqrt{7} \sqrt{3} - \sqrt{7} \sqrt{7}$ $sqrt2sqrt3-sqrt2sqrt7+sqrt7sqrt3-sqrt7sqrt7$ =

$\sqrt{6} - \sqrt{14} + \sqrt{21} - 7$ $sqrt6-sqrt14+sqrt21-7$

How do you multiply [√2+√7]⋅[√3−√7][sqrt(2) + sqrt(7)]*[sqrt(3) - sqrt(7)]?