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What are the real or imaginary solutions of the polynomial equation?
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\(64x^3+27=0\)
largely depends on the book
for first x=3/4
okay :) I know how to solve the second one when I plug it into the quadratic formula. Thank you! I just needed help getting started :)
yeah!!!!!!!!!!!!!!!!!!!!!!
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lol what's got you all excited?? :P
sorry the first () should be (4x+3) so x=-3/4
Okay :) thanks!
(4x+3)*(16x^2-12x+9)
Thank you again :)
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Does this look set up correctly @AlexandervonHumboldt2?\[x=\frac{ -(-12) \pm \sqrt {(-12)^2-4(16)(9)}}{ 2(16) }\]
yes
Okay, great :)
How can I simplify \(\sqrt {432}*i\)?
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