Question

What are the possible number of positive real, negative real, and complex zeros of

f(x) = 4x3 + x2 + 10x - 14

Answer 1

rule of signs ...

how many times do the signs (the opertors) change?

Answer 2

3 ?@amistre64

Answer 3

one idea is to get rid of those xs, let x = 1

4 + 1 + 10 - 14
^
we get 1 sign change when x is a positive value

there is at most, 1 postive root

now lets x = -1, what would that look like?

Answer 4

when x = -1, odd powers give us a negative, even powers are not effected:

(-1)^3 = -1
(-1)^2 = 1

so we essentially take the setup and negate the odd powers

Answer 5

4 * -1^3-1^2+10*-1-14

Answer 6

My home school treacvher justr gives me the work, i have no idea what im doing.

Answer 7

4x3 + x2 + 10x - 14

-4 + 1 - 10 - 14
^ ^
2 sign changes when x is a negative value
since this is an even number, we have to
assume that there can be complex values
as well

Answer 8

1 sign change when x=1, there is 1 positve root

2 sign changes when x = -1. there are 2 possible negative roots

since complex roots come in 2s, there might be 0 negative roots, and 2 complex roots

Answer 9

Positive Real: 1
Negative Real: 0
Complex: 2
@amistre64

Answer 10

like this ?

Answer 11

close, but we cant say for sure that there are 0 or 2 negative roots; or 0 or 2 complex roots

Positive Real: 1
Negative Real: 0, or 2
Complex: 0, or 2