Question

What are all the real zeros of y = (x - 12)3 - 10?

Answer 1

Try to expand the function then factor.

Answer 2

A. \[x = \sqrt[3]{10 + 12}\]
B. \[x = \sqrt[3]{10} + 12\]
C. x = \[\sqrt[3]{-10} + 12\]
D. \[x = \sqrt[3]{12} -10\]

Answer 3

how do i expand it could you help?

Answer 4

Wait, nevr mind about that. You can try and test each choice. One of them will cause the entire expression to be zero.

Answer 5

how do i test them?

Answer 6

@jim_thompson5910 please help.

Answer 7

\[\Large y = (x - 12)^3 - 10\]

\[\Large 0 = (x - 12)^3 - 10\]

\[\Large 0+10 = (x - 12)^3 - 10+10\]

\[\Large 10 = (x - 12)^3\]

\[\Large (x - 12)^3 = 10 \]

See what to do from here?

Answer 8

yes. the answer has to equal zero or what ... like what numbers do i plug in

Answer 9

your next step is to take the cube root of both sides

Answer 10

\[\Large y = (x - 12)^3 - 10\]

\[\Large 0 = (x - 12)^3 - 10\]

\[\Large 0+10 = (x - 12)^3 - 10+10\]

\[\Large 10 = (x - 12)^3\]

\[\Large (x - 12)^3 = 10 \]

\[\Large \sqrt[3]{(x - 12)^3} = \sqrt[3]{10} \]

\[\Large x - 12 = \sqrt[3]{10} \]

then what?

Answer 11

subtract 12 ? im not sure.

Answer 12

actually add 12 to both sides

Answer 13

\[\Large y = (x - 12)^3 - 10\]

\[\Large 0 = (x - 12)^3 - 10\]

\[\Large 0+10 = (x - 12)^3 - 10+10\]

\[\Large 10 = (x - 12)^3\]

\[\Large (x - 12)^3 = 10 \]

\[\Large \sqrt[3]{(x - 12)^3} = \sqrt[3]{10} \]

\[\Large x - 12 = \sqrt[3]{10} \]

\[\Large x - 12+12 = \sqrt[3]{10}+12 \]

\[\Large x = \sqrt[3]{10}+12 \]

Answer 14

so to sum things up: you plug in y = 0, then you solve for x

Answer 15

to get \[\Large x = \sqrt[3]{10}+12 \]

Answer 16

okay i get it a little better. thank you Jim

Answer 17

np