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Algebra: Solving Equations by Factoring

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Algebra: Solving Equations by Factoring [#permalink] New post 02 Feb 2011, 14:24
Hi all, I'm going through The Official Guide for GMAT Review (12th Edition). Unfortunately for me, I didn't take any math courses beyond what was required to graduate from high school.

I've hit a wall with Algebra: Solving Equations by Factoring. I'm given the following example:

x^3 – 2x^2 + x = –5(x – 1)^2:

x^3 - 2x^2 + x + 5(x – 1)^2 = 0
x(x^2 + 2x + 1) + 5(x – 1)^2 = 0
x(x – 1)^2 + 5(x – 1)^2 = 0
(x + 5)(x – 1)^2 = 0
x + 5 = 0 or (x – 1)^2 = 0
x = –5 or x = 1

I understand the first step, moving all of the expressions to one side, with 0 on the other side. I understand the next step, factoring the X. After that, I'm clueless. How the heck do you get from x(x^2 + 2x + 1) + 5(x – 1)^2 = 0 to x(x – 1)^2 + 5(x – 1)^2 = 0 ?

I tried to work backwards and did (x-1)^2. This came out to x^2 - 2x + 1... NOT the (x^2 + 2x +1) in the previous step. What am I doing wrong? Why and how are they doing what they are doing? I'm also completely clueless about the next step.

If someone can please hold my hand and explain this to me (keeping in mind that I'm a complete math moron and roll my eyes whenever the books say "so we can deduce that..."), it would be much appreciated!
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Re: Algebra: Solving Equations by Factoring [#permalink] New post 02 Feb 2011, 20:52
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x^3- 2x^2 + x = - 5(x - 1)^2:

x^3 - 2x^2 + x + 5(x - 1)^2 = 0

Here, you take x common from the first 3 terms. You get:

x(x^2 - 2x + 1) + 5(x - 1)^2 = 0

Now, x^2 - 2x + 1 = (x - 1)^2 This is an algebraic identity which you should learn. Anyway you can easily deduce it as you already did when you got (x - 1)^2 = x^2 - 2x + 1 as you mentioned below

x(x - 1)^2 + 5(x - 1)^2 = 0

Here, (x - 1)^2 is taken common so you are left with x from the first term and 5 from the second one.

(x + 5)(x - 1)^2 = 0

Product of 2 factors is 0. It means at least one of them has to be 0. e.g. you say if x*y = 0, either x or y or both should be 0. Otherwise you will not get the product 0. Same logic with factors.

x + 5 = 0 or (x – 1)^2 = 0
x = –5 or x = 1

Piece of Advice: OG12 will not give you the required theory. It is good for practice problems only. Since your concepts are rusty, get some basic algebra book e.g. we explain basic Math fundamentals in our book 'Math Essentials'. Check that out or any other basic Math book first.
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Re: Algebra: Solving Equations by Factoring [#permalink] New post 05 Feb 2011, 19:37
VeritasPrepKarishma, thank you very much!

Yes, I started to realize that doing the OG was probably not the best idea, especially since I need a lot of help with the theories. WIth that said, a friend loaned me the Manhattan GMAT books. I've been going through the number properties section, and that algebraic identity came up in a problem, but they had never explained it anywhere in the book. Very frustrating. And hard to research since I don't know what it's called... I didn't even know it was an "algebraic identity" until you mentioned it.

Is there anywhere where I can find other similar algebraic identities? While it's possible to deduce them, it's taking me forever to figure out problems because I just don't know about them. For example, I ran into...

"If n is a positive integer, is n^3 - n divisible by 4 (data sufficiency question)?"

I tried to simplify this by turning it into...
n(n^2 - n)

From there, I was stumped. Only after looking at the answer, I found out that I could and should have turned n(n^2 - n) into n(n - 1)(n + 1), which I could rearrange to be (n - 1)n(n + 1)... bingo! They're consecutive numbers! Now it makes sense. But as far as that step of turning the original equation into n(n - 1)(n + 1), I'm not clear on the rules and conditions that allow me to do this.

If there's somewhere where I can find a list and explanation of such algebraic identities/tricks/whatever they're called, please let me know!
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Re: Algebra: Solving Equations by Factoring [#permalink] New post 06 Feb 2011, 06:13
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GSDster wrote:
VeritasPrepKarishma, thank you very much!

Yes, I started to realize that doing the OG was probably not the best idea, especially since I need a lot of help with the theories. WIth that said, a friend loaned me the Manhattan GMAT books. I've been going through the number properties section, and that algebraic identity came up in a problem, but they had never explained it anywhere in the book. Very frustrating. And hard to research since I don't know what it's called... I didn't even know it was an "algebraic identity" until you mentioned it.

Is there anywhere where I can find other similar algebraic identities? While it's possible to deduce them, it's taking me forever to figure out problems because I just don't know about them. For example, I ran into...

"If n is a positive integer, is n^3 - n divisible by 4 (data sufficiency question)?"

I tried to simplify this by turning it into...
n(n^2 - n)

From there, I was stumped. Only after looking at the answer, I found out that I could and should have turned n(n^2 - n) into n(n - 1)(n + 1), which I could rearrange to be (n - 1)n(n + 1)... bingo! They're consecutive numbers! Now it makes sense. But as far as that step of turning the original equation into n(n - 1)(n + 1), I'm not clear on the rules and conditions that allow me to do this.

If there's somewhere where I can find a list and explanation of such algebraic identities/tricks/whatever they're called, please let me know!


The identities and other algebra basics have been discussed in the Veritas Algebra book though if you are just looking for the identities you need to know, here they are:

1. (x + y)^2 = x^2 + y^2 + 2xy
2. (x - y)^2 = x^2 + y^2 - 2xy
3. x^2 - y^2 = (x + y)(x - y) (called difference of squares - most important one. You should be able to recognize that x^2 - 9 = (x + 3)(x - 3) etc)

and if you would like to know how they are obtained,
(x + y)^2 = (x+y)(x+y) = x^2 + xy + yx + y^2 = x^2 + y^2 + 2xy

(x - y)^2 = (x-y)(x-y) = x^2 - xy - yx + y^2 = x^2 + y^2 - 2xy

(x+y)(x-y) = x^2 - xy + yx - y^2 = x^2 - y^2
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Re: Algebra: Solving Equations by Factoring [#permalink] New post 08 Feb 2011, 10:52
VeritasPrepKarishma wrote:
GSDster wrote:
VeritasPrepKarishma, thank you very much!

Yes, I started to realize that doing the OG was probably not the best idea, especially since I need a lot of help with the theories. WIth that said, a friend loaned me the Manhattan GMAT books. I've been going through the number properties section, and that algebraic identity came up in a problem, but they had never explained it anywhere in the book. Very frustrating. And hard to research since I don't know what it's called... I didn't even know it was an "algebraic identity" until you mentioned it.

Is there anywhere where I can find other similar algebraic identities? While it's possible to deduce them, it's taking me forever to figure out problems because I just don't know about them. For example, I ran into...

"If n is a positive integer, is n^3 - n divisible by 4 (data sufficiency question)?"

I tried to simplify this by turning it into...
n(n^2 - n)

From there, I was stumped. Only after looking at the answer, I found out that I could and should have turned n(n^2 - n) into n(n - 1)(n + 1), which I could rearrange to be (n - 1)n(n + 1)... bingo! They're consecutive numbers! Now it makes sense. But as far as that step of turning the original equation into n(n - 1)(n + 1), I'm not clear on the rules and conditions that allow me to do this.

If there's somewhere where I can find a list and explanation of such algebraic identities/tricks/whatever they're called, please let me know!


The identities and other algebra basics have been discussed in the Veritas Algebra book though if you are just looking for the identities you need to know, here they are:

1. (x + y)^2 = x^2 + y^2 + 2xy
2. (x - y)^2 = x^2 + y^2 - 2xy
3. x^2 - y^2 = (x + y)(x - y) (called difference of squares - most important one. You should be able to recognize that x^2 - 9 = (x + 3)(x - 3) etc)

and if you would like to know how they are obtained,
(x + y)^2 = (x+y)(x+y) = x^2 + xy + yx + y^2 = x^2 + y^2 + 2xy

(x - y)^2 = (x-y)(x-y) = x^2 - xy - yx + y^2 = x^2 + y^2 - 2xy

(x+y)(x-y) = x^2 - xy + yx - y^2 = x^2 - y^2


Thanks Karishma, you've been a huge help!

Regarding my original problem, is it safe to say that it was an error in the OG? Your explanation seems to confirm that x(x^2 + 2x + 1) + 5(x – 1)^2 = 0 really should have been x(x^2 - 2x + 1) + 5(x – 1)^2 = 0.

Is there an errata list somewhere?

EDIT: Not sure why, but whenever I try to use the math symbols tag, it turns my subtraction signs into a mess.
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Re: Algebra: Solving Equations by Factoring [#permalink] New post 08 Feb 2011, 12:04
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x^3 – 2x^2 + x = –5(x – 1)^2:

x^3 - 2x^2 + x + 5(x – 1)^2 = 0

Till this step, you have a '-' sign which is fine. I don't know how you got a '+' in the next step. When you take x common, the '-' sign stays as it is... Perhaps a typo at your end or in OG...

x(x^2 + 2x + 1) + 5(x – 1)^2 = 0

It should be x(x^2 - 2x + 1) + 5(x – 1)^2 = 0
It is fine after this.

x(x – 1)^2 + 5(x – 1)^2 = 0
(x + 5)(x – 1)^2 = 0
x + 5 = 0 or (x – 1)^2 = 0


P.S. - Use a hyphen for '-' sign. It will stay put.
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Re: Algebra: Solving Equations by Factoring [#permalink] New post 08 Feb 2011, 12:19
VeritasPrepKarishma wrote:
x^3 – 2x^2 + x = –5(x – 1)^2:

x^3 - 2x^2 + x + 5(x – 1)^2 = 0

Till this step, you have a '-' sign which is fine. I don't know how you got a '+' in the next step. When you take x common, the '-' sign stays as it is... Perhaps a typo at your end or in OG...

x(x^2 + 2x + 1) + 5(x – 1)^2 = 0

It should be x(x^2 - 2x + 1) + 5(x – 1)^2 = 0
It is fine after this.

x(x – 1)^2 + 5(x – 1)^2 = 0
(x + 5)(x – 1)^2 = 0
x + 5 = 0 or (x – 1)^2 = 0


P.S. - Use a hyphen for '-' sign. It will stay put.


Yes, I double-checked the OG and they went from x^3 - 2x^2 + x + 5(x-1)^2 = 0 to x(x^2 + 2x + 1)+4(x-1)^2=0. So, I guess it was a typo on their end. So frustrating. :roll:

I sort of figured out the problem I was having with the math tags... apparently, copy and pasting equations into here can mess things up. I retyped the equations manually and they worked fine.

Thanks again for all your help!
Re: Algebra: Solving Equations by Factoring   [#permalink] 08 Feb 2011, 12:19
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