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# Factorisation

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Intern
Joined: 10 Apr 2012
Posts: 47
Concentration: Finance
Schools: Goizueta '19 (I)
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Kudos [?]: 22 [0], given: 13

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07 Jun 2012, 01:29

Is it not (x+3)(x+1)(x-1)

I don't seem to get why its x(x + 3)(x – 1)
Intern
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07 Jun 2012, 01:35
1
KUDOS
x^3 + 2x^2 = 3x

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

x ( x^2 + 2x -3 ) = 0

x ( x^2 + 3x - x - 3 ) = 0

x [ x(x+3) - 1(x+3) ] = 0

x(x+3)(x-1) = 0
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07 Jun 2012, 21:49
1
KUDOS
Cosmas wrote:

Is it not (x+3)(x+1)(x-1)

I don't seem to get why its x(x + 3)(x – 1)

Do it step by step

x^3 + 2x^2 = 3x

transfer 3x to the other side we have

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

we have a common factor of x among the three, as such we have

x (x^2 + 2x - 3) = 0

Now we can still factor x + 2x - 3 into

(x + 3) (x - 1)

if you are having a hard time to believe me, let's use FOIL:

x*x + 3x - 3 - 1*x = x^2 + 2x - 3

Got it?

That's why we have:

x (x + 3) (x - 1)
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07 Jun 2012, 23:13

x^3 + 2x^2 = 3x
x^3 + 2x^2 -3x = 0
x * (x^2 + 2x - 3) = 0
x * (x+3) * (x-1) = 0

Voila. I've always struggled with factorization as well. Usually once I get to the quadratic stage, I just do trial and error (i.e., from x^2 + 2x - 3 to (x+3)(x-1)). Is there a quicker method?
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07 Jun 2012, 23:28
I get it now. Thanks buddies! I figured that i have to factorise (x^2 + 2x - 3) as well.

gmatsaga wrote:
Cosmas wrote:

Is it not (x+3)(x+1)(x-1)

I don't seem to get why its x(x + 3)(x – 1)

Do it step by step

x^3 + 2x^2 = 3x

transfer 3x to the other side we have

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

we have a common factor of x among the three, as such we have

x (x^2 + 2x - 3) = 0

Now we can still factor x + 2x - 3 into

(x + 3) (x - 1)

if you are having a hard time to believe me, let's use FOIL:

x*x + 3x - 3 - 1*x = x^2 + 2x - 3

Got it?

That's why we have:

x (x + 3) (x - 1)
Re: Factorisation   [#permalink] 07 Jun 2012, 23:28
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