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znimon
 2 years ago
If I have x^5 81x and I want to find all the zeros here is how I go about that: x(x^481), x(x^43^4), x((x^2)^2(3^2)^2), x((x^23^2)(x^2+3^2), x((x3)(x+3)(x+3i)(x3i))
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.
znimon
 2 years ago
If I have x^5 81x and I want to find all the zeros here is how I go about that: x(x^481), x(x^43^4), x((x^2)^2(3^2)^2), x((x^23^2)(x^2+3^2), x((x3)(x+3)(x+3i)(x3i)) 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.

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hartnn
 2 years ago
Best ResponseYou've already chosen the best response.11(3*10) = 30 it cannot be +30 distribution happens in addition/subtraction to multiplication, like a(b+c) = ac+ac

znimon
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks a ton, could you expand a little bit if possible on how that relates to the polynomial above.

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1what you have done is correct x^581x= x((x3)(x+3)(x+3i)(x3i)) so zeros are 0,3,3,3i,3i

znimon
 2 years ago
Best ResponseYou've already chosen the best response.0I think I am confused because there is addition, subtraction, and multiplication inside the parentheses in the above polynomial.

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1all the steps are correct, which step u have doubt in particular ?

znimon
 2 years ago
Best ResponseYou've already chosen the best response.0The need for all the parenthesis in the final step.

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1yes, that paranthesis is needed x(x^4+81) must be there.

znimon
 2 years ago
Best ResponseYou've already chosen the best response.0Will you expand on that a little. I still feel stuck for some reason.

znimon
 2 years ago
Best ResponseYou've already chosen the best response.0Its not quite clear. I understand the difference now between 1(3+10) and 1(3*10)

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1x^581x= x.x^481x Taking x common =x(x^481) if u don't put paranthesis , u will have x.x^481 which is x^581 which is incorrect

znimon
 2 years ago
Best ResponseYou've already chosen the best response.0Oh thanks that make a lot of sense.

hartnn
 2 years ago
Best ResponseYou've already chosen the best response.1glad i could clear it :)
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