Question

how to rationalize an expression???

Answer 1

depends on the expression

Answer 2

\[\sqrt[3]{32y^9z ^{14}}\]

Answer 3

one part is easy
since \(3\) goes in to \(9\) \(3\) times, the \(y^9\) comes out of the radical as \(y^3\)

Answer 4

since \(32=2^5\) you have
\[\large \sqrt[3]{2^5}=2\sqrt[3]{2}\]

Answer 5

so do you take the \[y^3\] out front?

Answer 6

yes, the \(y^3\) comes out front

Answer 7

as for the \[\sqrt[3]{z^{14}}\] the easiest way to think about it is this:

3 goes in to 14 4 times, with a remainder of 2

out comes \(z^4\) in stays \(z^2\)

Answer 8

you good with that or no?

Answer 9

that's confusing!

Answer 10

ok lets do it another way

Answer 11

\[\sqrt[3]{z^{14}}=\sqrt[3]{z^3\times z^3\times z^3\times z^3\times z^2}\]
\[=z\times z\times z\times z \times \sqrt[3]{z^2}\]
\[=z^4\sqrt[3]{z^2}\]

Answer 12

but that is because when you add the exponents \(3+3+3+3+2=14\) in other words \(4\times 3+2=14\)

Answer 13

okay! i get it now!

Answer 14

that is why it is easier to think
"3 goes in to 14 four times, with a remainder of two"
out comes \(z^4\) in stays \(z^2\)
it is easier to think about it that way

Answer 15

so the answer is technically \[2y^3z^4\sqrt[3]{2z^2}\]

Answer 16

technically and actually, yes

Answer 17

nice use of equation editor btw

Answer 18

thank you! :)

Answer 19

oh wait!! mistake !!

Answer 20

\[2y^3z^4\sqrt[3]{2^2z^2}\]

Answer 21

i made a mistake earlier
\[\large \sqrt[3]{2^5}=2\sqrt[3]{2^2}\]

Answer 22

also i would like to add that this is not called "rationalizing" it is called "writing in simplest radical form"

Answer 23

would you like to help me simplify one more? @satellite73