Question

Simplifying radical expressions

Answer 1

ok now we're working with third roots instead of square roots....
so instead of looking for PAIRS of factors in the radicand, you'll be looking for triple factors....

let's start by factoring 32 until you can't break it down any further.....

Answer 2

32 = ????

Answer 3

give me a sec.

Answer 4

one thousand one...

Answer 5

hahaha. ok. i thik 2*16?

Answer 6

keep breaking down whatever factor you can... so break down that 16....

Answer 7

ok. so 8..then..4, then 2?

Answer 8

ok... so write out the prime factorization of 32...

32 =

Answer 9

32=2*2*8?

Answer 10

that 8 can be broken down... keep going....

Answer 11

32=2*2*4*8?

Answer 12

no

Answer 13

ok what did I do wrong then?

Answer 14

2*2*2*4?

Answer 15

keep going... that 4 can be broken down....

Answer 16

2*2*2*2?

Answer 17

do you get 32 when you multiply 2*2*2*2 ????

Answer 18

no i get 16. so its 2*2*2*2*2? if not, then im confused as to what it should be

Answer 19

yes...
32 = 2 * 2 * 2 * 2 * 2

this is what is meant by PRIME factorization.. when all the factors are prime numbers....

Answer 20

ok i see that now..thankyou!

Answer 21

ok so back to the problem...

Answer 22

\(\large \sqrt[3]{32}=\sqrt[3]{2 \cdot 2 \cdot2 \cdot2 \cdot 2} \)

how many triple 2's do you see inside the radical sign?

Answer 23

1

Answer 24

good... how many left over?

Answer 25

2

Answer 26

ok... so cross out those three 2's inside the radical and write it once OUTSIDE:
\(\large \sqrt[3]{32}=2\sqrt[3]{\cancel {2 \cdot 2 \cdot2} \cdot2 \cdot 2}=2\sqrt[3]{2 \cdot 2}=2\sqrt[3]{4} \)

Answer 27

so its \[\sqrt[3]{32}=2\sqrt[3]{2\times2}\]

Answer 28

\[=2\sqrt[3]{4}\]

Answer 29

yes

Answer 30

ok, i see that i have one more..and another one, but Im going to try to do that one on my own. could you help me on one more?

Answer 31

sure... new post pls...

Answer 32

will do. thank you