Question

If you have for Descartes Rules of Sign 1 positive zero and 2 or 0 negative roots. What is your complex root?

Answer 1

The Descartes rule of signs gives you the potential number of positive, negative and complex roots. It doesn't give the specific roots.

Answer 2

Are you asking what is the number of potential complex roots?

Answer 3

If so, what degree is the polynomial?

Answer 4

3

Answer 5

It is f(x)=x^3-x^2-6x-70

Answer 6

Ok, you have a 3rd degree polynomial that has 3 roots.
By Descartes, you have a potential 1 positive root and a potential 2 or 0 negative roots.

Answer 7

Yes

Answer 8

So this is really a find the roots problem...

Answer 9

Remember that if you have a polynomial with real corefficients, if there are complex roots, they come is pairs of two complex conjugate roots.

Answer 10

Now you can use rational root theorem to suss out your real root. It will be one of the factors (positive or negative) of 70.

Answer 11

Possibilities:
1 positive root, 2 negative roots
1 positive root, 2 complex roots

Answer 12

So there are 3 zeros and that means there are 3 complex roots?

Answer 13

3 complex roots is not an option in a polynomial with only real coefficients.

Answer 14

No. You can't have 3 complex roots bec complex roots must come in multiples of 2.

Answer 15

they always come in pairs: \(a \pm bi\) where \(i = \sqrt{-1}

Answer 16

ackthpth. \(i = \sqrt{-1}\)

Answer 17

Ok let's start small. Because the Fundamental Theorem says there are 3 total zeros

Answer 18

So how many complex zeros are there?

Answer 19

That's good.
1. There are 3 roots in total.
2. The roots are potentialy:1 positive root and 2 or zero negatice roots.
Ok?

Answer 20

Ok

Answer 21

That's good.
1. There are 3 roots in total.
2. The roots are potentialy: 1 positive root and 2 or zero negative roots.
Ok?

Answer 22

So how do I know how many complex roots are there?

Answer 23

Remember, complex roots must come in pairs. You can't have just one copmplex root.

Answer 24

With this in mind, let's go over the possibilities of what the roots may be.

Answer 25

1 positive root, 2 negative roots, 0 complex roots
1 positive root, 0 negative roots, 2 complex roots

Answer 26

So 2 complex roots

Answer 27

Or is it 2 or 0 complex roots

Answer 28

We don't know. Descartes tells us the potential roots, not definite.
We definitely will have 1 positive root.
The other two will be either two negative roots or two complex roots.

Answer 29

So how do we find out?

Answer 30

The reason 1 positive root is definite, is that there is no other way of figuring it out, once you put together the potential negative of complex roots.

Answer 31

The only way to find out is to find the actual roots.
Since one root is positive, start looking for it as a factor of 70, using synthetic division.

Answer 32

Once you find the positive root, then you divide the polynomial to get a quadratic. Then you can use the quadratic formula to find the other two roots.

Answer 33

-5 is a root

Answer 34

I turned it to 5 and used synthetic

Answer 35

and got no remainders making it a zero

Answer 36

-5 is not a root, but 5 is.

Answer 37

Yeah sorry

Answer 38

Good.
Now what did you get as the quotient of the synthetic division? That is the new quadratic polynomial that you need to use the quadratic formula with.

Answer 39

I got x^2+4x-14

Answer 40

Good.
Now use the quadratic formula.

Answer 41

x=2.24 x=−6.24

Answer 42

Can't be. You got 1 positive and 1 negative roots of the quadratic.
That would make 2 positive and 1 negative roots in total.
That's not one of our possibilities.
The roots of the quadratic must be either both negative or both complex.

Answer 43

−2±32√2

Answer 44

and 2+-32sq2

Answer 45

Sorry. I looked quickly and didn't notice your mistake above. The quadratic is
x^2 + 4x + 14, not - 14

Answer 46

-(-x+3 sqrt(2)-2) (x+3 sqrt(2)+2)

Answer 47

Is that it?