Question

what is the simplified form of (sqrt)128n17

Answer 1

\[\sqrt{128n^{17}}=\sqrt{2*64*n*n^{16}}\]\[=8n^8\sqrt{2n}\]

Answer 2

Do you understand what I am doing to get this answer?

Answer 3

you find the factors of 128

Answer 4

correct and i know that \(\sqrt{64}=8\) and \(\sqrt{a^n}=(a^n)^{1/2}=a^{n/2}\)

Answer 5

So I take out everything I can that gives a whole number answer and what is left I leave in the square root sign

Answer 6

okay so if i have 5(sqrt)5 + 3 (sqrt) 20 - 6 (sqrt)3 i would fins the factors of 5 ,20 and 3

Answer 7

find

Answer 8

yes but because 5 and 3 are prime you can only find the factors of 20

Answer 9

which is 4 times 5 which would lead to 2 (times)2(times)5?

Answer 10

right so it would be \(3 \sqrt{20}=3*\sqrt{2}*\sqrt{2}*\sqrt{5}=3*2*\sqrt{5}=6\sqrt{5}\)

Answer 11

okay so putting that in to the expression would you get 11(sqrt)5-6(sqrt)3?

Answer 12

perfect. :D

Answer 13

yayy thanks