Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

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

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

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

Offline

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

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.

Offline

**irspow****Member**- Registered: 2005-11-24
- Posts: 457

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.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

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.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Actually the katy's polynomial is unractorizable.

IPBLE: Increasing Performance By Lowering Expectations.

Offline

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

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)*

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

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.

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

```
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**

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Yes. Other roots are immagineric.

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

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.

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

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

**igloo** **myrtilles** **fourmis**

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

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.

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

|-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**

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

The coeficient for x^2:

(x-a1)(x-a2)(x-a3) =>-a3x^2-a2x^2-a1x^2

-(a1+a2+a3)x^2=**+3**x^2 =>

a1+a2+a3=**-3**

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Oops, my mistake. Nice explaining**!**

**igloo** **myrtilles** **fourmis**

Offline

Pages: **1**