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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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hartnnBest ResponseYou've already chosen the best response.1
1(3*10) = 30 it cannot be +30 distribution happens in addition/subtraction to multiplication, like a(b+c) = ac+ac
 one year ago

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

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

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

hartnnBest ResponseYou've already chosen the best response.1
all the steps are correct, which step u have doubt in particular ?
 one year ago

znimonBest ResponseYou've already chosen the best response.0
The need for all the parenthesis in the final step.
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
yes, that paranthesis is needed x(x^4+81) must be there.
 one year ago

znimonBest ResponseYou've already chosen the best response.0
Will you expand on that a little. I still feel stuck for some reason.
 one year ago

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

hartnnBest ResponseYou've already chosen the best response.1
x^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
 one year ago

znimonBest ResponseYou've already chosen the best response.0
Oh thanks that make a lot of sense.
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
glad i could clear it :)
 one year ago
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