Question

Use Descartes' rule of signs to determine the number of possible positive, negative, and nonreal complex solutions of the equation.
4x3 − 8x2 + x − 4 = 0

A)possible positive: 2 or 0
possible negative: 1 or 0
possible complex: 0 or 3

B)possible positive: 1 or 0
possible negative: 3 or 1
possible complex: 0 or 2

C)possible positive: 3 or 1
possible negative: 0 or 2
possible complex: 0

D)possible positive: 3 or 1
possible negative: 0
possible complex: 0 or 2

E)possible positive: 4 or 2
possible negative: 0
possible complex: 0 or 2

whats the answer..?

Answer 1

Well, count! What do you get?

Answer 2

this is familiar to me but i cant really remember..

Answer 3

Count the sign changes. That's it.

Answer 4

is it 3..?

Answer 5

That's what I get. This rules out all but C and D.

Find f(-x) and do it again. (Really, just change the sign on all the odd exponents.)

Answer 6

is it C..?

Answer 7

Hmmm... You seem to be guessing.

f(x) = 4x3 − 8x2 + x − 4 That's 3 sign changes, so 3 or 1 positive Real.
f(-x) = -4x3 − 8x2 - x − 4 That's 0 sign changes, so 0 negative Real.
What does that leave for Complex? Looks like 0 or 2.

Answer 8

yeah.. i cant really remember how.. even im counting sign changes.. hehe.. sorry..

Answer 9

oh.. now i got it.. haha.. tnx.. i remembered it again.. haha

Answer 10

but wait.. what does possible complex means..?

Answer 11

With Real coefficients, any Complex roots must appear in conjugate pairs. If you can't account for them amongst the Reals, they must be Complex.

Answer 12

wow... i didnt get it.. haha

Answer 13

Your only possibilities are:

3 positive real, 0 negative real, 0 Complex
1 positive real, 0 negative real, 2 Complex

See how those two POSSIBLE positive Reals jumped over to Complex?

Answer 14

ahhh.. ok.. i got it now..