If I have x^5 -81x and I want to find all the zeros here is how I go about that: x(x^4-81), x(x^4-3^4), x((x^2)^2-(3^2)^2), x((x^2-3^2)(x^2+3^2), x((x-3)(x+3)(x+3i)(x-3i))
I am confused as whether the parentheses separating the (alone) x from the rest of the equation are necessary. For example -1(3*10), distribute the negative one is +30, but if I do what is in the parentheses first I get -30. I know the associative property and cumulative property as I understand them.

See more answers at brainly.com

Thanks a ton, could you expand a little bit if possible on how that relates to the polynomial above.

what you have done is correct
x^5-81x= x((x-3)(x+3)(x+3i)(x-3i))
so zeros are 0,3,-3,3i,-3i

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.