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Factoring Complex Numbers
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Default Factoring Complex Numbers - 10-28-2011, 09:08 AM

Yea, I should know this probably lol. But, I don't. Never been good with this.

I need help factoring the following equations.

1) 2x^4 - 4x^2 + 10x
2) 6x^4 - 2x^3 - 3x^2 + x
3) 6x^2 - 23x + 7

I remember my teacher talking about a 'slide and divide' method.. but I can't remember how to do it :3





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Default 10-28-2011, 09:25 AM

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Originally Posted by Ries View Post
1) 2x^4 - 4x^2 + 10x
2) 6x^4 - 2x^3 - 3x^2 + x
3) 6x^2 - 23x + 7
1) First, seeing as they all have x in common, we factor that out.
x(2x^3-4x+10)
Not sure it goes any further.

2)What we do here is separate the equation.
(6x^4 - 2x^3) + (-3x^2 + x)
Now factor.
2x^3(3x - 1) - x(3x - 1)
As you can see they have (3x - 1) in common.
So the answer is;
(2x^3 - x)(3x - 1)

3)Ok. For this one, we multiply 6 by 7.
6x^2 - 23x + 7
It'll be 42. From that 42, we get two numbers that multiplied will equal 42 and added will equal -23. The numbers are -21 and -2.
6x^2 -2x -21x + 7
Group up the numbers like we did in #2 and factor.
(6x^2 -2x) + (-21x +7)
2x(3x - 1) -7 (3x - 1)

We see they have (3x - 1) in common. SO the answer is;
(2x - 7)(3x - 1)


If there is something here you didn't understand, let me know.
I will try to explain it better.




Last edited by Mr 223; 10-28-2011 at 09:31 AM.
  
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Default 10-28-2011, 09:36 AM

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Originally Posted by Mr 223 View Post


1) First, seeing as they all have x in common, we factor that out.
x(2x^3-4x+10)
Not sure it goes any further.

2)What we do here is separate the equation.
(6x^4 - 2x^3) + (-3x^2 + x)
Now factor.
2x^3(3x - 1) - x(3x - 1)
As you can see they have (3x - 1) in common.
So the answer is;
(2x^3 - x)(3x - 1)

3)Ok. For this one, we multiply 6 by 7.
6x^2 - 23x + 7
It'll be 42. From that 42, we get two numbers that multiplied will equal 42 and added will equal -23. The numbers are -21 and -2.
6x^2 -2x -21x + 7
Group up the numbers like we did in #2 and factor.
(6x^2 -2x) + (-21x +7)
2x(3x - 1) -7 (3x - 1)

We see they have (3x - 1) in common. SO the answer is;
(2x - 7)(3x - 1)


If there is something here you didn't understand, let me know.
I will try to explain it better.
Thanks (:
You're the best <3

EDIT: Any suggestions on 'x^3 + 125'?





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Default 10-28-2011, 09:49 AM

Quote:
Originally Posted by Ries View Post


Thanks (:
You're the best <3

EDIT: Any suggestions on 'x^3 + 125'?
Basically we are using this. a^3 + b^3.
The answer to this is; (a + b)(a^2 - ab + b^2)
NOTE: If it was a^3 - b^3, the answer changes completely.

So, the cube root of x^3 is x.
For 125, it is 5. (5 x 5= 25 x 5=125)
(X + 5)( X^2 -5X +25)
Can't do anything from there.




Last edited by Mr 223; 10-28-2011 at 09:53 AM.
  
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Default 10-28-2011, 10:09 AM

1) 2x (x^3 - x + 10)
2) x (6x^3 - 2x^2 - 3x + 1)
I didn't really know this one, so i just went with this, although it's kinda dumb...
3) 1 ( 6x^2 - 23x + 7)

I think what ur teacher means by 'slide and divide' is that you have to check for a common factor (by sliding and checking them 1 by 1) for example, the common factor between 3x, 4x and 23x is x. and the one between 12x 4x^3 and 6x^2 is 2x
  
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