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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^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.

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  1. hartnn
    • 2 years ago
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    -1(3*10) = -30 it cannot be +30 distribution happens in addition/subtraction to multiplication, like a(b+c) = ac+ac

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

  3. hartnn
    • 2 years ago
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    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

  4. znimon
    • 2 years ago
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    I think I am confused because there is addition, subtraction, and multiplication inside the parentheses in the above polynomial.

  5. hartnn
    • 2 years ago
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    all the steps are correct, which step u have doubt in particular ?

  6. znimon
    • 2 years ago
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    The need for all the parenthesis in the final step.

  7. hartnn
    • 2 years ago
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    yes, that paranthesis is needed x(x^4+81) must be there.

  8. znimon
    • 2 years ago
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    Will you expand on that a little. I still feel stuck for some reason.

  9. znimon
    • 2 years ago
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    Its not quite clear. I understand the difference now between -1(3+10) and -1(3*10)

  10. znimon
    • 2 years ago
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    @hartnn

  11. hartnn
    • 2 years ago
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    x^5-81x= x.x^4-81x Taking x common =x(x^4-81) if u don't put paranthesis , u will have x.x^4-81 which is x^5-81 which is incorrect

  12. znimon
    • 2 years ago
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    Oh thanks that make a lot of sense.

  13. hartnn
    • 2 years ago
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    glad i could clear it :)

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