Question

Part 1 : What are the possible number of positive, negative, and complex zeros of f(x) = 2x3 – 5x2 – 6x + 4 ?
Part 2 : Use complete sentences to explain the method used to solve this equation.

Answer 1

@Luis_Rivera

Answer 2

@precal

Answer 3

@Zarkon

Answer 4

HINT: Descartes Rule.

Answer 5

descartes rule confuses me. you dont need to give me the answer to this question. i would just like some explanation as to how to simplify it and such

Answer 6

@FoolForMath

Answer 7

i did

Answer 8

Descartes' Rule of Signs is a useful help for finding the zeroes of a polynomial, assuming that you don't have the graph to look at. This topic isn't so useful if you have access to a graphing calculator because, rather than having to do guess-n-check to find the zeroes (using the Rational Root Test, Descartes' Rule of Signs, synthetic division, and other tools), you can just look at the picture on the screen. But if you need to use it, the Rule is actually quite simple.

Answer 9

@FoolForMath

Answer 10

Which is the intended approach? Descartes rule of graphing calculator?

Answer 11

i dont have one..
and to look for which ones are positive and which ones are negatives, right?

Answer 12

@Answerthisnow

Answer 13

How many change of sign are there in f(x) ?

Answer 14

3?

Answer 15

@FoolForMath

Answer 16

o.o

Answer 17

@LagrangeSon678

Answer 18

1. positive zeroes for counting the sign changes for f(x)
2. negative zeroes for counting the sign changes for f(-x) = 2x^3 - 5x^2 - 6x + 4.

Good Luck!

Answer 19

http://www.wolframalpha.com/input/?i=graph+f%28x%29+%3D+2x3+%E2%80%93+5x2+%E2%80%93+6x+%2B+4+

Answer 20

sorry bad graph, hard to draw with a mouse

Answer 21

three