Question

Find a cubic function with the given zeros:
sqrt2, -sqrt2, -2 :)

Answer 1

general form of cubic eq.-
\[x^{3}-(\alpha +\beta+ \gamma)x^{2}+(\alpha \beta+\beta \gamma+\alpha \gamma)x-\alpha \beta \gamma=0\]where alpha beta and gamma are the zeroes.

Substitute the values of zeroes.

Answer 2

or why dont you just multiply out
(x-a)(x-b)(x-c) where a,b,c are roots..

Answer 3

same thing

Answer 4

(x-sqrt2)(x+sqrt2)(x+2)?

Answer 5

\[x^{3}-(\sqrt{2}-\sqrt{2}-2)x^{2}+(-\sqrt{2}\sqrt{2}+2\sqrt{2}-2\sqrt{2})x-2\sqrt{2}\sqrt{2}=0\]
\[x^{3}-(-2)x^{2}+(-2)x-2*2=0\]
\[x^{3}+2x^{2}-2x-4=0\]

Answer 6

So, x^2-2 (x+2)
x^3 +2x^2 -2x -4?

Answer 7

Oh. Yeah!