## General Question # How many possible positive and negative zeros are in 3x^5 + 6x?

Asked by mathhelp1221 (11 ) October 14th, 2014 from iPhone

I have a question about Descartes’ rule of signs, and I can’t figure out how many possible positive and negative zeros there are in 3x^5 + 6x. Can somebody help? Thanks!

Observing members: 0 Composing members: 0  You do not need the rule of signs to determine the number of positive and negative roots. Start by factoring out x to get x(3x^4 + 6) = 0. We have one root at x=0. Let’s check for the solutions of 3x^4 + 6 = 0. Rearranging the terms gives x^4 = -2. This has no real solutions, positive or negative, since any real number raised to an even power must be greater than or equal to 0.

LostInParadise (27572 )“Great Answer” (6 ) Flag as…  To use the sign rule, notice that the two coefficients have the same sign (3 and 6 are both positive), so there is no sign change in f(x), which means that there are no positive real roots. To work with the negative real roots, find f(-x) = -3x^5 – 6x. Again there are no sign changes, so there are no negative real roots.

LostInParadise (27572 )“Great Answer” (2 ) Flag as… or