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#1 2005-12-30 06:03:19

katy
Member
Registered: 2005-12-28
Posts: 14

Factor Theorem?

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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#2 2005-12-30 07:46:19

Flowers4Carlos
Member
Registered: 2005-08-25
Posts: 106

Re: Factor Theorem?

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.

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#3 2005-12-30 07:47:29

irspow
Member
Registered: 2005-11-24
Posts: 455

Re: Factor Theorem?

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.

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#4 2005-12-30 08:50:55

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Factor Theorem?

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


IPBLE:  Increasing Performance By Lowering Expectations.

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#5 2005-12-30 08:51:57

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Factor Theorem?

Actually the katy's polynomial is unractorizable.


IPBLE:  Increasing Performance By Lowering Expectations.

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#6 2005-12-30 09:47:30

deepu
Member
Registered: 2005-12-30
Posts: 3

Re: Factor Theorem?

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

Last edited by deepu (2005-12-30 09:49:26)

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#7 2005-12-30 10:04:26

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Factor Theorem?

It has one zero but it isn't rational:

Last edited by krassi_holmz (2005-12-30 10:05:41)


IPBLE:  Increasing Performance By Lowering Expectations.

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#8 2005-12-30 11:41:53

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

Re: Factor Theorem?

                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.

Last edited by John E. Franklin (2005-12-30 11:43:55)


igloo myrtilles fourmis

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#9 2005-12-30 23:43:39

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Factor Theorem?

Yes. Other roots are immagineric.


IPBLE:  Increasing Performance By Lowering Expectations.

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#10 2005-12-31 00:02:44

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Factor Theorem?

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


IPBLE:  Increasing Performance By Lowering Expectations.

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#11 2005-12-31 03:37:49

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

Re: Factor Theorem?

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


igloo myrtilles fourmis

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#12 2005-12-31 03:46:17

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Factor Theorem?

Here is it:
(x-a1)(x-a2)(x-a3)=-a1 a2 a3 + (a1 a2 + a1 a3 + a2 a3) x - (a1 + a2 + a3) x² + x³.
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)


IPBLE:  Increasing Performance By Lowering Expectations.

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#13 2005-12-31 19:37:11

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

Re: Factor Theorem?

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


igloo myrtilles fourmis

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#14 2005-12-31 21:40:39

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Factor Theorem?

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


IPBLE:  Increasing Performance By Lowering Expectations.

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#15 2006-01-01 04:34:12

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,562

Re: Factor Theorem?

Oops, my mistake.  Nice explaining!


igloo myrtilles fourmis

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