Question

Need algebra 2 help! See Attachment!

Answer 1

first divide 810 by 2
what do you get?

Answer 2

you get 405

Answer 3

so we are at \[\sqrt[4]{405}\] as a first step
now we need to factor \(405\) to see if it contains any fourth powers in its factored form

Answer 4

wait find a factor of 405?

Answer 5

do you know what i mean by that?

Answer 6

not quite :/

Answer 7

ok, lets to it
it is pretty clear that \(405\) is divisible by 5 right?

Answer 8

Yes

Answer 9

so since \(405\div5=81\) we know \[405=5\times 81\]

Answer 10

that is what i mean by factoring 405

Answer 11

ohh okay I see so far.

Answer 12

now we are not done we can do one of two things now, depending on how good our multiplication tables are
we can continue factoring 81 or we can recognize that \(81=9\times 9\)

Answer 13

and since \(9=3\times 3\)this means \[81=3\times 3\times 3\times 3=3^4\]

Answer 14

so there was a fourth power lurking inside \(405\) we have \[405=3^4\times 5\] in factored form

Answer 15

Ohh okay so you sort of took a shortcut or shortened it.

Answer 16

that makes \[\sqrt[4]{405}=\sqrt[4]{3^4\times 5}\]which in turn is \[\sqrt[4]{3^4}\sqrt[4]{5}\]

Answer 17

Making 81 to \[3^4\]

Answer 18

yeah i took a short cut to do that

Answer 19

otherwise you keep dividing by 3 until you get it

Answer 20

what about the 4 and the squareroot, those stay?

Answer 21

now we have \[\sqrt[4]{3^4}\times \sqrt[4]{5}\] as a second to last step

Answer 22

and it should be pretty clear that \(\sqrt[4]{3^4}=3\)

Answer 23

Yes

Answer 24

so final answer is \[3\sqrt[4]{5}\]

Answer 25

that is called "simplest radical form"

Answer 26

Okay makes total sense now, thank you!

Answer 27

yw