Math Is Fun Forum

phamthephong100

Member

Registered: 2007-10-13

Posts: 8

Who can solve this problem???

NullRoot

Member

Registered: 2007-11-19

Posts: 162

Re: Who can solve this problem???

It factors out to:
(2x - 1)(x²  - 2x + 3)

X can be 0.5 so that (2x-1) = 0

For (x²  - 2x + 3) = 0, you need to go into complex numbers. I think the answer's something like 1 ± i√2

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Trillian: Five to one against and falling. Four to one against and falling… Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you still can’t cope with is therefore your own problem.

luca-deltodesco

Member

Registered: 2006-05-05

Posts: 1,470

Re: Who can solve this problem???

x²  - 2x + 3 = (x-1)²+2
x-1 = ±i√2
x = 1±i√2

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The Beginning Of All Things To End.
The End Of All Things To Come.

NullRoot

Member

Registered: 2007-11-19

Posts: 162

Re: Who can solve this problem???

Thanks, Luca. I thought that was right, but wasn't sure. I knew there weren't any real answers, though

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Trillian: Five to one against and falling. Four to one against and falling… Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you still can’t cope with is therefore your own problem.

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Re: Who can solve this problem???

After looking at the equation:

2x^3-5x^2+8x-3=0

How would one know that it factors out to:

2x - 1)(x²  - 2x + 3)

I mean, after having the factors, it is obvious but starting with the equation what are the steps to figure out the factors?

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Who can solve this problem???

Here's one way...

There can be one linear and one quadratic factor, so

So we are left with

Solving these simultaneously gives

,

,

.

So

In fact, that's the only way I can think of, unless you are a genius with algebraic manipulation or you get lucky with the factor theorem.