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Precalculus 16 Online
OpenStudy (anonymous):

Find a cubic function with the given zeros. You must show your work for credit. -2, 3, -3

OpenStudy (anonymous):

There are multiple cubic functions that would work.

OpenStudy (anonymous):

like what because this problem is really getting me confused

OpenStudy (anonymous):

Given the roots \(r_1, r_2, r_3\) a factored cubic equation would look like: \[ (x-r_1)(x-r_2)(x-r_3)=0 \]

OpenStudy (anonymous):

So, we can just use the function (x + 2)(x - 3)(x + 3) = y. (x - 3)(x + 3) = x^2 - 9 (x + 2)(x^2 - 9) = x^3 + 2x^2 - 9x - 18 So, y = x^3 + 2x^2 - 9x - 18

OpenStudy (anonymous):

it makes since now thank you . so how do you do it when square roots are involved

OpenStudy (anonymous):

How do you do what?

OpenStudy (anonymous):

find the cubic function like square root of 2 , square root of -2 , and 2

OpenStudy (anonymous):

Do you mean negative square root of two rather than square root of negative two? If not, then you are dealing with imaginary numbers, which is by all means doable (just not in this specific circumstance), but very different from what you are doing now.

OpenStudy (anonymous):

yes i meant negative square root

OpenStudy (anonymous):

Oh good. So, you would just expand (x + sqrt(2)) * (x - sqrt(2)) * (x - 2) Which is (x^2 - 2)(x - 2) = x^3 - 2x^2 - 2x + 4

OpenStudy (anonymous):

so that will be the y right ?

OpenStudy (anonymous):

yes. y = x^3 - 2x^2 - 2x + 4

OpenStudy (anonymous):

thank you now i have a much better understanding :)

OpenStudy (anonymous):

You're welcome! Glad to help!

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