two actually really hard problems. probably only top contributors can figure it out. please help?

Question

Explain how the use of Descartes' Rule of Signs can simplify the process of finding the zeros of the following polynomial:

P(x) = 12x^4 + x^3 + 4x^2 + 7x + 8

Using the theorems and techniques of this section, find the zeros of the following polynomial:

P(x) = x^3 - 4x^2 + x + 6

Answer 1

These aren't hard at all.

First one:

P(x) = 12x^4 + x^3 + 4x^2 + 7x + 8

4th power, therefore four zeros, all real, all imaginary, or two of each. Zero sign changes, therefore zero positive real roots.

P(-x) = 12x^4 - x^3 + 4x^2 - 7x + 8

4 sign changes, therefore 0, 2 or 4 negative real zeros.

Now you do the second.

Answer 2

Hi,
1) Explain how the use of Descartes' Rule of Signs can simplify the process of finding the zeros of the following polynomial:
P(x) = 12x^4 + x^3 + 4x^2 + 7x + 8

Descartes' Rule of signs can tell you that there are no positive or negative real roots for this polynomial. Descartes rule says that the number of positive real rational roots is equal to the number of sign changes in the function or less than that by an even number.
For negative roots, the rule applies similarly if -x is substituted for x in the function.

2) Using the theorems and techniques of this section, find the zeros of the following polynomial:
P(x) = x^3 - 4x^2 + x + 6
Since my telepathic powers are rather weak these days :-), I don't know what theorems and techniques you had in "this section," in fact I don't have a clue where "this section" is located. But I'll use a few common techniques.

P(x) = x^3 - 4x^2 + x + 6
First of all there are two sign changes, so there are either two positive real roots or zero.
P(-x) =-x^3-4x² -x +6
There is one sign change, so there is one negative real root.
Now, the possible real roots are +- the factors of 6 divided by 1. So, that's +- {1, 2, 3, 6}
Just by inspection you can see that with -1 substituted in the polynomial, you'll have -1 -4-1+6 = 0. So, let's do synthetic division using -1 to find the reduced equation:

__-1__| 1.. -4...1... 6
.............__-1__5__-6
,,,,,,,,,,,,,1...-5...6....0
So, -1 is a root and we have the reduced equation, and we can find the other roots by factoring the reduced equation.
x²-5x+6 =0
(x-3)(x-2)=0
x-3=0
x =3
x-2=0
x=2
So, we have {-1,2, 3} as roots.

FE

Answer 3

Descartes Rule will not give the exact zeros, but only tell about possible positive or negative real roots or zeros by counting changes in the sign of successive terms of polynomial.If there are n sign changes, the possible number of zeros are, n or (n-2) or (n-4).........upto zero. This is termed as counting by pairs


P(x) = 12x^4 + x^3 + 4x^2 + 7x + 8

Here all the signs are + only, so there is no change of sign. So there is no real positive zero

replace x with -x
P(-x) = +12x^4 - x^3 + 4x^2 - 7x + 8

Now sign has changed alternately. So there are 4 sign changes.

There may be 4 or 2 or zero negative real roots

2)

P(x) = + x^3 - 4x^2 + x + 6

here there two sign changes between x^3 and -4x^2 and between -4x^2 and +x

so there may be 2 or zero positive real roots

P(-x) = -x^3 - 4x^2 - x + 6

there is only one sign change. So there is exactly one negative root.

P(-1) = -1 - 4 - 1 + 6 = -6 + 6 = 0

so -1 is one of the zeros.

P(2) = 8 - 16 + 2 + 6 = 16 - 16 = 0

2 is another zero

p(3) = 27 - 36 + 3 + 6 = 36 - 36 = 0

so all the zeros of x^3 - 4x^2 + x + 6 are

-1, 2 and 3

Answer 4

The Descartes rule of signs can tell how many factors are negative and how many are positive , by counting how many times the equation switches signs.

Take the first equation :

P(x) = 12x^4 + x^3 + 4x^2 + 7x + 8

There's no change in sign in P(x) so it means that there's no positive root.

Then P (-x) =12x^4 - x^3 + 4x^2 - 7x + 8

The equation witched signs 4 times (from 2 positive-negative , 2negative-positive) . Therefore there are 4 negative roots.



2) Use the rule of signs ( 2 positive roots and 1 negative root)

p/q= 6/1 .
List all possible factors of p/1 = +-1,+-2,+-3,+-6

Synthetic division:

x^3 - 4x^2 + x + 6

1-4+1+6 |3

__3_-3__-6_____
1-1 -2 + 0

So we get one factor: x=3

We use the reduced equation which is

X^2-x - 2=0

(X-2)(x+1)

So the three zeroes of the euqation would be>

X= 2, X=-1 , x=3. 2 positive roots and one negative root.

more questions?
Check this site out:

http://www.ask4mytutor.com

they give free trial tutoring