Question

Part 1 [3 points]: What are the possible number of positive, negative, and complex zeros of f(x) = -2x3 - 5x2 - 6x + 4 ?
Part 2 [4 points]: Use complete sentences to explain the method used to solve this equation.

Answer 1

@rock_mit182

Answer 2

@iambatman

Answer 3

anyone know?

Answer 4

use the rational roots thrm ....

Answer 5

I don't know how to do that

Answer 6

then you need to review your material .... you have to have some idea as to what it is in order to be able to iron out any wrinkles you may have about it.

Answer 7

if you can give me a sensible explanantion of the rational roots thrm, ill be able to guide you in applying it here.

Answer 8

my online school doesn't provide help to any of this its a learn on your own kinda thing and im completely confused

Answer 9

heres the thrm then ...

given a polynomial with constant coefficients (what is a coefficient?)

if it has any rational roots, they must come from a pool of options such that the factors (what are factors) of the last term, divided by the factors of the first term form the possible roots

Answer 10

\[P(x)=ax^n+bx^{n-1}+...+k\]
if this has any rational roots, they must be of the form:\[\pm\frac{<a>}{<k>}\]

Answer 11

such that <j> is the set of all the factors of j

Answer 12

but re reading it tells me the the sign test is more appropriate to finding the number of +-complex roots

i dont spose your familiar with the sign test?

Answer 13

no im not

Answer 14

the way the signs .. those +- parts of the equation .... the way they change tells us something about the positive, negative, and complex roots

Answer 15

the simplest thing to do is to let x=1 .. a positive number to get postive roots

-2x3 - 5x2 - 6x + 4 ?

-2 - 5 - 6 + 4
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0 1

the signs change only once, this tells us that there is at most, 1 postive root

Answer 16

let x=-1 to compare negative roots

-2(-1)3 - 5(-1)2 - 6(-1) + 4 ?

+2 - 5 +6 + 4 ?
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0 1 2

the signs change twice, there are at most 2 but possibly 0 negative roots

the possibility for 0 negative is that the complex roots come in pairs ... sets of 2 that can replace a set of 2+ or 2-

so there are possibly 2 or 0 complex roots

Answer 17

since we know there is at least 1 positive root, it will have to be one of the rational roots from the pool of options:

factors of 4 are 1,2,4
----
factors of 2 are 1,2


the pool of options are: 1/1, 2/1, 4/1, 1/2, 2/2, 4/2

or simplified as:{1, 2, 4, 1/2} then we just trial and error it.