anonymous
  • anonymous
T or F any 3rd degree polynomial has at at least one real root?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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cherylim23
  • cherylim23
True, can you explain why?
anonymous
  • anonymous
True
ash2326
  • ash2326
Yeah @timo86m , but the third degree polynomial should have real coefficients

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anonymous
  • anonymous
yes i can @cherylim23 because x^2 has u or n shape but 3rd degree has a |dw:1342235273380:dw| shape
cherylim23
  • cherylim23
Condition i s that all coefficients are real. Do you know why? This relates back to the fundamental theorem of algebra, which states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. However, if the coeffients are all real, you can apply the complex conjugate theorem here, leaving you definitely one real root at least since all complex roots would have occured in conjugate pairs. Bit hard to understand if you haven't learnt complex numbers yet, but just keep this in mind
anonymous
  • anonymous
i know basic complex numbers like a HS level knowledge of them. But I know if it is all real then a shape like |dw:1342235527228:dw| should always intersect the x axis at least once. in a u or n shaped graph due to a 2nd dagree polynomial there are ways to shift the curve up or down so that it never intersects the x axis.
anonymous
  • anonymous
another reason is that the complex roots come in conjugate pairs, so you cannot have just one of them or three of them
anonymous
  • anonymous
good point

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