What are the possible numbers of positive real, negative real, and complex zeros of f(x) = -7x4 - 12x3 + 9x2 - 17x + 3? Positive Real: 3 or 1 Negative Real: 1 Complex: 2 or 0 Positive Real: 3 or 1 Negative Real: 2 or 0 Complex: 1 Positive Real: 1 Negative Real: 3 or 1 Complex: 2 or 0 Positive Real: 4, 2 or 0 Negative Real: 1 Complex: 0 or 1 or 3
@goformit100
like this ? \(\large f(x) = -7x4 - 12x^3 + 9x^2 - 17x + 3\)
thought of rational root thm ?
yes thats right and im having trouble wrapping my head around it
oklet me see ^^
does the rational roots test solve it?
i guess yes lets try
but i think its for real mmm idk abt complex
i got plus or minus 1, 1/7, 3, 3/7
so we need to find p,q \(\Huge q=\pm 1,\pm3\) \(\Huge p=\pm 1,\pm7\) so possible zeros \(\Huge \frac{p}{q} =\pm 1,\pm3 ,\pm\frac{3}{7}, \pm \frac{1}{7} \)
ok lol
okay whats the next step?
mmm nw we need to check from equation ?
well for rational its done
we can eliminate d and c
also we can remove b
thank you(:
nop :D
?? Might be wandering off on this one. We are not asked to FIND the roots, only to COUNT them. What are the possible numbers of positive real, negative real, and complex zeros of f(x) = -7x4 - 12x3 + 9x2 - 17x + 3 Descartes Rule of Signs suggests that we count the sign changes of f(x) --+-+ That's 3 changes. There are, therefore, 3 or 1 Positive Real Roots. Descartes Rule of Signs suggests that we count the sign changes of f(-x) -++++ That's 1 change. There must be, therefore, 1 Negative Real Root. Since there are 4 total, there can be 3 Positive, 1 Negative, and 0 Complex or 1 Positive, 1 Negative, and 2 Complex.
mmm it said what are not how many i guess
I agree that "possible numbers" could go either way. There we have both, in any case.
ic ... i liked ur method though
Never underestimate the power of something that can count! Always a good thing to know.
yeah !
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