Answer: A
Step-by-step explanation:
[tex](\sqrt{12}+\sqrt{6})(\sqrt{6}-\sqrt{10})[/tex]
[tex](\sqrt{12}*\sqrt{6})-(\sqrt{12}* \sqrt{10})+(\sqrt{6}*\sqrt{6})-(\sqrt{6}*\sqrt{10})[/tex]
[tex](\sqrt{12*6})-(\sqrt{12*10})+(\sqrt{6*6})-(\sqrt{6*10})[/tex]
Let's rewrite the numbers inside to see if we can take some out.
For example; 12 is 3 x 4 and 4 is [tex]2^2[/tex] which means a 2 will come out of the root.
[tex](\sqrt{3*2^2*6})-(\sqrt{3*2^2*10})+(\sqrt{6^2})-(\sqrt{6*10})[/tex]
[tex](2\sqrt{18})-(2\sqrt{30})+(6)-(\sqrt{60})[/tex]
We can also rewrite 18 as 9 x 2 and 9 can be rewritten as [tex]3^2[/tex] which will make the 3 come out.
60 can be rewritten as 15 x 4 and 4 can be rewritten as [tex]2^2[/tex] which will make the 2 come out.
Let's rewrite this.
[tex](2\sqrt{2*3^2})-(2\sqrt{30})+(6)-(\sqrt{15*2^2})[/tex]
When a number comes out of a root, it has to be multiplied by the number that is already outside.
[tex](2*3\sqrt{2})-(2\sqrt{30})+(6)-(2\sqrt{15})[/tex]
Solve and eliminate parentheses.
[tex]6\sqrt{2}-2\sqrt{30}+6-2\sqrt{15}[/tex]
