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Factoring

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Current Student
Joined: 03 Dec 2007
Posts: 84
Schools: NYU STERN
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Factoring [#permalink]  18 Jan 2009, 12:20
I am having trouble factoring problems that having something in front of the x^2

For example- how would I do this?

3x^2 - 8x -3 = 0

OG says (3x+1)(x-3)

But how do you figure that out quickly?!
Manager
Joined: 15 Apr 2008
Posts: 166
Followers: 2

Kudos [?]: 10 [0], given: 1

Re: Factoring [#permalink]  18 Jan 2009, 13:19
hardaway7 wrote:
I am having trouble factoring problems that having something in front of the x^2

For example- how would I do this?

3x^2 - 8x -3 = 0

OG says (3x+1)(x-3)

But how do you figure that out quickly?!

this is a quadratic equation. here's how you solve it.
you have to multiply the number before x^2 and the constant. so in this eq. 3 *3=9

now you factorize 9 in such a way that when you add or substract the factors you get -8x. the factors of 9 are 1, 3 and 9.
the equation can be simplified as follows

3x^2 -9x + 1x -3=0
3x(x-3) + 1(x-3)=0
(3x+1)(x-3)=0

i hope it is clear.
Current Student
Joined: 03 Dec 2007
Posts: 84
Schools: NYU STERN
Followers: 3

Kudos [?]: 6 [0], given: 0

Re: Factoring [#permalink]  18 Jan 2009, 13:35
Thanks ALD-

How did you go from: 3x(x-3) + 1(x-3)=0

to:
(3x+1)(x-3)=0

Sorry I am slow!!! I don't know what is wrong with me!
Director
Joined: 12 Jul 2008
Posts: 518
Schools: Wharton
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Re: Factoring [#permalink]  18 Jan 2009, 13:39
1
KUDOS
hardaway7 wrote:
Thanks ALD-

How did you go from: 3x(x-3) + 1(x-3)=0

to:
(3x+1)(x-3)=0

Sorry I am slow!!! I don't know what is wrong with me!

now the two components of the sum -- 3x(x-3) and 1(x-3) -- both have (x-3) in common. So you can factor out (x-3) from each term.

Last edited by zonk on 18 Jan 2009, 13:40, edited 1 time in total.
Director
Joined: 14 Oct 2007
Posts: 759
Location: Oxford
Schools: Oxford'10
Followers: 13

Kudos [?]: 183 [1] , given: 8

Re: Factoring [#permalink]  18 Jan 2009, 13:39
1
KUDOS
hardaway7 wrote:
Thanks ALD-

How did you go from: 3x(x-3) + 1(x-3)=0

to:
(3x+1)(x-3)=0

Sorry I am slow!!! I don't know what is wrong with me!

Hi Hardaway7,

if you look at 3x(x-3) + 1(x-3),
(x-3) is common to the parts being added together. So one can factor this out.

its the same as 3ab + 1b can be written as b(3a + 1)

good luck

-buff
Senior Manager
Joined: 25 Nov 2006
Posts: 337
Schools: St Gallen, Cambridge, HEC Montreal
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Kudos [?]: 36 [0], given: 0

Re: Factoring [#permalink]  18 Jan 2009, 13:55
What are doing here, Buff?
Are you re-taking the GMAT like me?
Current Student
Joined: 03 Dec 2007
Posts: 84
Schools: NYU STERN
Followers: 3

Kudos [?]: 6 [0], given: 0

Re: Factoring [#permalink]  18 Jan 2009, 13:58
ahhh got it!

THANKS to both of you!
Director
Joined: 14 Oct 2007
Posts: 759
Location: Oxford
Schools: Oxford'10
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Kudos [?]: 183 [0], given: 8

Re: Factoring [#permalink]  18 Jan 2009, 13:59
lumone wrote:
What are doing here, Buff?
Are you re-taking the GMAT like me?

Hi Lumone,

truth be told, being a fellow martian, I was compelled to help Hardaway7.

back to the application forums!

-buff
Manager
Joined: 04 Jan 2009
Posts: 243
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Re: Factoring [#permalink]  19 Jan 2009, 10:03
hardaway7 wrote:
I am having trouble factoring problems that having something in front of the x^2

For example- how would I do this?

3x^2 - 8x -3 = 0

OG says (3x+1)(x-3)

But how do you figure that out quickly?!

a fail proof way to solve a problem of the type
ax^2+bx+c=0 is
x=-b+/-sqrt(b^2-4ac)

A simpler approach is to look at the actual values of ac and b (with the signs).
Then find two integers k1 and k2 such that k1+k2 = ac and k1k2=b.
In the present situation,
ac=-9 and b=-8. Thus, k1=-9 and k2 = 1
and we get
3x^2-9x+x-3=0 and problem solved.
_________________

-----------------------
tusharvk

Re: Factoring   [#permalink] 19 Jan 2009, 10:03
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