how do u determine the number of positive negative and complex zeros in f(x)=4x^3 -3x^2 +10x -5
@satellite73
hello again
hi im in a big hurry and have 2 problems to do
you want the zeros or the possible zeros?
the zeros im pretty sure
there is only one zero and it is positive but you cannot find it i think they want you to use descartes rule of sign
there are 3 changes in sign of the coefficients, so there are either positive zeros and no complex zerso, or one positive zero and 2 complex zeros in fact there is one positive and two complex
i am not sure what your answer choices are
also note that \[f(-x)=-4x^3-3x^2-10x-5\] so there are NO possible negative zeros
there are no choices it just says demonstrate how to determine the number of positive negative and complex zeros
ok say that \[f(x)=4x^3 -3x^2 +10x -5 \] has three changes in sign, from \(+4\) to \(-3\), from \(-3\) to \(+10\) and from \(+10\) to \(-5\) therefore there are either 3 positive zeros and no complex zeros (since there are at most thee zeros) or 1 positive zero and two complex zeros, since you count down by twos
then say \[f(-x)=-4x^3-3x^2-10x-5\] has NO changes in sign, so there cannot be any negative zeros
conclusion 1 positive, two complex OR 3 positive no complex in fact there is only one positive and two complex zeros, but i don't think it is asking you the actual answer, just the how to find the possible ones
thank you soo much
yw on line class?
yes i have one more can you please help
k
write a polynomial function using the roots 1, -2 and 3i
how quick you want it?
pretty quick lol
i mean you want the method or the answer ?
awnser please
junt the funtion not solved
ok then we multiply \[(x-1)(x+2)(x^2+9)\] using wolfram
and get \[x^4+x^3+7 x^2+9 x-18\]
so that would be the function?
\[f(x)=x^4+x^3+7 x^2+9 x-18\]
yes, that is it
thank u
yw
you done?
yes thank youu
whew take a break from math glad to help
wait sorry one more thing :)
what are the steps to solving tht
wait nvm got it
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