f(x) = 2x^4 +4x^3-5x^2-5x+6 how many zeros of the function must be imaginary

Question

consider the graph

Answer 1

Answer 2

max 2 positive real zeros, & max 2 negative real zeros given Descartes' Rule of Signs

either 0, 2 or 4 imaginary zeros

if the graph intersects the x axis 4 times, there are no imaginary zeros

if it intersects 2 times, there are 2 imaginary zeros

if it never intersects, there are 4 imaginary zeros

if it just touches the x axis once (tangent but not crossing) that counts the same as 2 intersections, then there are 2 imaginary zeros

if it just touches the x axis twice, that counts as 4 intersections and then there are no imaginary zeros

x =about -1.39 and -2.39 are the two zeros

or about -1.4 and -2.4

or about -7/5 and -12/5

x f(x)

0 6

1 2+4-5-5+6 = 2

2 32+32-40-10+6 =8

no change of signs for positve x, so there are no postive zeros

-1 2-4-5-5+6 =4

-2

the change in sign of f(x) as x changes from -1 to -2 means there is a zero somewhere between -1 and -2 such as -1.39

-2 32-32 -20+10+6= -4

-3 162-108-45+15+6=+36

another change in sign from x=-2 to -3. a zero is between them: -2.39

use a graphing calculator

it intesects the x axis twice left of the y axis, so there are 2 negative real zeros. 0 positive real zeros and 2 imaginary zeros

the 2 zeros are about -1.39 and -2.39

the graph is "W" shaped with only the left half of the W intersectng the x axis.to the left of the y axis

imaginary zeros come in conjugate pairs so there has to be 0,2 or 4 imaginary zeros. Turns out there are 2