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

Question

anonymous · Answer

attachment

anonymous · Answer

@nathan917

anonymous · Answer

@rajee_sam

anonymous · Answer

@Numb3r1

anonymous · Answer

If you plug each of those into the equation, you'll see which ones produce zero.

anonymous · Answer

Frankly, only one of them makes sense.

anonymous · Answer

However, there will probably be other zeros than those listed. For that, I'll expand it, but take a look at those first and see which one works.

rajee_sam · Answer

$$(x-12)^{3} -10 = 0$$$$(x-12)^{3} =10$$$$(x-12) = \sqrt[3]{10}$$$$x = \sqrt[3]{10} + 12$$

anonymous · Answer

I tried A and it worked it was =0

anonymous · Answer

Now, there still may be other zeros from more subtle factorizations, but if you are only asked which of those options is a zero of the function, A is correct.

anonymous · Answer

$$(x-12)^3+10=x^3-36x^2+432x-1728+10=x^3-36x^2+432x-1718$$ I don't know if that was all you were asked for, but if there are more roots they'll be roots of this equation.

anonymous · Answer

That's the only root. Tested and true.