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Topic review (newest first)

John E. Franklin
2006-01-02 03:34:12

Oops, my mistake.  Nice explaining!

krassi_holmz
2006-01-01 20:40:39

The coeficient for x^2:
(x-a1)(x-a2)(x-a3) =>-a3x^2-a2x^2-a1x^2
-(a1+a2+a3)x^2=+3x^2 =>
a1+a2+a3=-3

John E. Franklin
2006-01-01 18:37:11

|-a1a2a3=-15
|a1a2+a2a3+a3a1=13
|a1+a2+a3=-3   <------shouldn't that be positive 3 ?
Also, thanks for the explanation, I see now.

krassi_holmz
2006-01-01 02:46:17

Here is it:
(x-a1)(x-a2)(x-a3)=-a1 a2 a3 + (a1 a2 + a1 a3 + a2 a3) x - (a1 + a2 + a3) x + x.
The coeficients for the same powers of x must be same, so:
a1 a2 a3 = 15 AND
a1 a2 + a1 a3 + a2 a3 = 13 AND
a1 + a2 + a3 = -3

(
|equation1
|equation2
...
|equationm
means equation1 && equation2 && ... equationm
(system)

John E. Franklin
2006-01-01 02:37:49

How did you get |a1a2+a2a3+a3a1=13 ?? and what does the | symbol mean?

krassi_holmz
2005-12-31 23:02:44

A factorization in complex numbers:
x^3+3x^2+13x-15=(x-a1)(x-a2)(x-a3)
So
|-a1a2a3=-15
|a1a2+a2a3+a3a1=13
|a1+a2+a3=-3

krassi_holmz
2005-12-31 22:43:39

Yes. Other roots are immagineric.

John E. Franklin
2005-12-31 10:41:53

Code:

                x^3 + 3x^2 + 13x - 15
                the factor theorm, what is that?
                x (x^2 + 3x + 13) = 15
                Guess 1, too big
                Guess .9,  .9 (.81 + .27 + 13) = 15
                Guess .95, too big
                Guess .92,  15.28
                Guess .91,  15.07
                Guess .9067, 14.999 Close enough.
                So factor out (x - .9067)

                x - .9067       x^3 + 3x^2 + 13x - 15

      quotient:       x^2 + (3 + .9067)x  + (13 + .9067(3 - .9067))   There is a remainder so imaginary pair??

 dividing calculations:  x^3  -  .9067x^2    -    .9067(3 - .9067)x   -  .9067(13 + .9067(3 - .9067))

      krassi_holmz's mathimatica got: .906756576

Because it didn't divide almost evenly, instead of getting -15 on end, it multiplies to about -13.5,
I guess this means the other roots aren't real numbers??  I'm just guessing from what
krassi said.

krassi_holmz
2005-12-31 09:04:26

It has one zero but it isn't rational:

deepu
2005-12-31 08:47:30

Hi Katy

According to the factor theorem,
For a polynomial P(x), x - a is a factor if P(a) = 0.

therefore substituting x=1 in the polynomial
P(x) =(x^3+3x^2+13x-15)

P(1) = 1+3+13-15
        =2

again substituting x=2 in P(x)

P(2)= 8+12+26-15
       =31

P(3)=27+27+39-15
       =93

As the remainder is not zero for x=1,2,3,4,.............so on

Hence the given polynomial P(x) =(x^3+3x^2+13x-15)
has no factor

I think the polynomial u have given is unfactorizable

krassi_holmz
2005-12-31 07:51:57

Actually the katy's polynomial is unractorizable.

krassi_holmz
2005-12-31 07:50:55

irspow wrote:

If you divide your equation by x-1 you will get x² - 2x - 15

   This means that (x-1)( x² - 2x - 15) = x³ + 3x² +13x - 15

   Factoring the second part of the product gives (x-5)(x+3)

   So (x-1)(x-5)(x+3) =  x³ + 3x² +13x - 15

   This is completely factored because there are no powers of x above one.

Not.
(x-1)(x^2-2x-15)=x^3-2x^2-15x-x^2+2x+15=x^3-3x^2-13x+15

irspow
2005-12-31 06:47:29

If you divide your equation by x-1 you will get x² - 2x - 15

   This means that (x-1)( x² - 2x - 15) = x³ + 3x² +13x - 15

   Factoring the second part of the product gives (x-5)(x+3)

   So (x-1)(x-5)(x+3) =  x³ + 3x² +13x - 15

   This is completely factored because there are no powers of x above one.

Flowers4Carlos
2005-12-31 06:46:19

hello katy!

aren't you given a lovely divisor like (x-c) to try out??  otherwise we need to figure out where this function equals to zero and that requires using the rational zeros theorem.

katy
2005-12-31 05:03:19

I am stuck with this peoblem...someone please help!!

Completely factor (x^3+3x^2+13x-15) using the factor theorem.

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