Question

I need help multiplying these radicals:) there’s two: A) and B)

Answer 1

Answer:

Step-by-step explanation:

Well lets start with

(3*sqrt 2+ 1)(2*sqrt 3-4)

Lets multiply everything in the second parenthesis by 3*sqrt 2

2 sqrt 3 * 3 sqrt 2 =  6 sqrt 6

-4*3 sqrt 2 =  -12 sqrt 2

Now lets multiply everything by 1

1*2 sqrt 3 = 2 sqrt 3

1*-4 = -4

we have

-4  -12 $\sqrt{2}$  +  6 $\sqrt{6}$  + 2 $\sqrt{3}$  as the awnser to problem A

Now lets do problem b

We can start by multiplying everything in the second parenthesis by 2

2*5=10

2*-1 *sqrt 3 = -2 sqrt 3

Now multiply everything in the second parenthesis by 2sqrt3

2sqrt 3 * 5 = 10 sqrt 3

2 sqrt 3 * -1* sqrt3 =  -6

Our final awnser is

10-6 +10 sqrt 3 -2 sqrt 3 ->  4+ 8 $\sqrt{3}$

The awnser to question B is 4+ 8 $\sqrt{3}$

pls give brainliest