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cherylim23
 2 years ago
Best ResponseYou've already chosen the best response.0True, can you explain why?

ash2326
 2 years ago
Best ResponseYou've already chosen the best response.2Yeah @timo86m , but the third degree polynomial should have real coefficients

timo86m
 2 years ago
Best ResponseYou've already chosen the best response.0yes i can @cherylim23 because x^2 has u or n shape but 3rd degree has a dw:1342235273380:dw shape

cherylim23
 2 years ago
Best ResponseYou've already chosen the best response.0Condition i s that all coefficients are real. Do you know why? This relates back to the fundamental theorem of algebra, which states that every nonconstant singlevariable 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

timo86m
 2 years ago
Best ResponseYou've already chosen the best response.0i 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.

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0another reason is that the complex roots come in conjugate pairs, so you cannot have just one of them or three of them
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