Question

really need help on this i dont know where to start.
i will give a medal to whoever helps

Answer 1

You need to factor 75, 8, and 32 into their prime factorizations.
Do you know how to do that?

Answer 2

no idea:/

Answer 3

The prime numbers are 2, 3, 5, 7, 11, 13, etc.
Start with 8.
Can you write 8 as a product of prime numbers?

Answer 4

no because none of those two mulitply to 8

Answer 5

You can use the same prime number more than once.

Answer 6

you could use 2 x 2 x 2

Answer 7

For example:
Write 27 as a product of primes.
Answer: 27 = 3 * 3 * 3 = 3^3

Answer 8

Ok, here is another approach.
Are you familiar with the perfect squares, 1, 4, 9, 16, 25, 36, 49, etc.?
They are the squares of 1, 2, 3, 4, 5, 6, 7, etc.

Answer 9

ok so 8 would be 2^3

Answer 10

Great. Exactly.

Answer 11

What about 75?

Answer 12

25 times 3?

Answer 13

Good.
25 is also 5 * 5, so you can also say 5^2 * 3

Answer 14

What about 32?
Can you break down 32 into a product of prime numbers?

Answer 15

2^5?

Answer 16

yes 2^5 = 32

for the square roots, look for pairs of the same number and take them out of the square root but put only one of the pair on the outside:
example
sqrt(3*2*2) = 2*sqrt(3)
(we "took out" 2*2 , and put a single 2 out front)
or
sqrt(3*3*3*3)
here there are 2 pairs of 3's, so 3 sqrt(3*3) = 3*3 sqrt(1) (you are left with sqrt(1) if you take everything out) or just 3*3 or 9
so sqrt(81)= 9 (as a calculator will show)

Answer 17

so the sqr(75) is sqr(5*5*3)
there is one pair of 5's that you can take out, (and put only 1 5 out front)
\[ \sqrt{75} = \sqrt{5\cdot 5 \cdot 3} = 5 \sqrt{3} \]
can you try the other two terms ?