BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

factoring strategies?

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 17
Joined: Tue Aug 05, 2008 4:15 pm

factoring strategies?

by penumbra547 » Mon Feb 09, 2009 9:20 pm
hi,

besides the hit-and-miss method, how would one know which to factor or distribute in order to solve the following equation:

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

I instinctively try to take care of the (x-1)^2 just because i feel like sometimes there are opportunities to cancel variables out. but other than that, i really don't have a strategy.

anyone have a more methodical way of attacking this?

thanks in advance.
Top
Source: — Quantitative Reasoning |
User avatar
Site Admin
Posts: 2567
Joined: Thu Jan 01, 2009 10:05 am
Thanked: 712 times
Followed by:550 members
GMAT Score:770

by DanaJ » Tue Feb 10, 2009 1:10 am
x^3-2x^2+x = x(x^2 -2x + 1) = x(x-1)^2. Now, this is equal to -5(x - 1)^2, which means that x(x - 1)^2 + 5(x - 1)^2 = 0 or that (x + 5)(x - 1)^2. This has two solutions:
x = -5
x = 1

Hope this helps.
Top
Junior | Next Rank: 30 Posts
Posts: 17
Joined: Tue Aug 05, 2008 4:15 pm

by penumbra547 » Tue Feb 10, 2009 9:14 am
DanaJ,

Thanks for the response. While I arrived at the answer, I'm just curious what made you think to factor out the "x" on the left side of the equation first, versus FOILing out the right side of the equation first?

What tipped you off?

I believe the same answer is achieved regardless of which side you approach first, but I believe starting from the right side of the equation takes longer to arrive at the answer. And I want to save as much time as possible on the test.

Thanks!
Top
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Tue Aug 26, 2008 6:20 pm
Location: new york

Quick Question

by dsheppard » Tue Feb 10, 2009 10:06 am
Please explain how you get from x(x - 1)^2 + 5(x - 1)^2 = 0 TO (x + 5)(x - 1)^2.

Thank You

x^3-2x^2+x = x(x^2 -2x + 1) = x(x-1)^2. Now, this is equal to -5(x - 1)^2, which means that x(x - 1)^2 + 5(x - 1)^2 = 0 or that (x + 5)(x - 1)^2. This has two solutions:
x = -5
x = 1
Top
Junior | Next Rank: 30 Posts
Posts: 17
Joined: Tue Aug 05, 2008 4:15 pm

by penumbra547 » Tue Feb 10, 2009 10:08 am
you get to (x + 5)(x - 1)^2=0 by factoring out (x-1)^2 from both sides of the equation.

equation=x(x - 1)^2 + 5(x - 1)^2
Last edited by penumbra547 on Tue Feb 10, 2009 10:12 am, edited 2 times in total.
Top
User avatar
Site Admin
Posts: 2567
Joined: Thu Jan 01, 2009 10:05 am
Thanked: 712 times
Followed by:550 members
GMAT Score:770

by DanaJ » Tue Feb 10, 2009 10:09 am
Well, there are two things you should keep in mind:
1. From what I can tell, GMAT is not about crazy third, fourth or fifth degree equations. The roughest it could get is a second degree equation with pretty obvious roots. So if you see equations with a degree greater than two, you should always consider factoring in order to "shed some weight" off the damn thing.

2. When you see smth like x^3-2x^2+x, you should immediately notice that there is no "free factor" (I'm not sure this is the correct English translation), meaning smth that is not "tied" to an x. This tipped me off in this particular case. For further clarifications, let me give you some examples:

a. Equations with "free factors" (again, I'm sorry if this is not the correct translation):
x^5 + 6x + 3 = 0 (that 3 is the "free factor")
x^2 - 5x + 6 = 0 (6 = "free factor")
20x^4 - 3x^3 + 7 = 0(7 = "free factor")

b. Equations with no "free factors":
x^5 + x^4 + 5x^2 = 0, which is equivalent to x^2(x^3 + x^2 + 5) = 0
x^3 - 4x^2 + 3x = 0, which is equivalent to x(x^2 - 4x + 3) = 0. The expression between the parenthesis is a pretty basic second degree equation: x^2 - 4x + 3 = (x - 3)(x - 1). This means that the initial equation is x(x - 3)(x - 1) = 0, with 3 solutions: 0, 3 and 1. This is the type of factoring you should expect in the GMAT.


Hope this clears up things.
Top
Junior | Next Rank: 30 Posts
Posts: 17
Joined: Tue Aug 05, 2008 4:15 pm

by penumbra547 » Tue Feb 10, 2009 10:16 am
danaJ,

thank you, for the colorful metaphors! I'll remember to 'shed some weight' and look for "free factors."

i never would have guessed english isn't your first language. where are you from?
Top
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Tue Aug 26, 2008 6:20 pm
Location: new york

Re:

by dsheppard » Tue Feb 10, 2009 10:17 am
Wow..Thank you for that explanation, It is very helpful and clears up more than just this problem.

My question is: How do you account for the 5 outside of 5(x - 1)^2 ? I would assume you distribute it but you are somehow creating an (x+5) from it? correct?

That is what confuses me?
Top
Junior | Next Rank: 30 Posts
Posts: 17
Joined: Tue Aug 05, 2008 4:15 pm

by penumbra547 » Tue Feb 10, 2009 10:22 am
from x(x - 1)^2 + 5(x - 1)^2=0

to

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

you simply factor out (x-1)^2 from x(x - 1)^2 + 5(x - 1)^2=0
Top
Junior | Next Rank: 30 Posts
Posts: 11
Joined: Tue Aug 26, 2008 6:20 pm
Location: new york

by dsheppard » Tue Feb 10, 2009 10:31 am
aahh...Thank you so much for your help!
Top
User avatar
Site Admin
Posts: 2567
Joined: Thu Jan 01, 2009 10:05 am
Thanked: 712 times
Followed by:550 members
GMAT Score:770

by DanaJ » Tue Feb 10, 2009 10:36 am
Let's start from the top and maybe I'll be able to shed some light on it this time. So, you start with:
x^3-2x^2+x=-5(x-1)^2
Now, we've already established that x^3 - 2x^2 + x = x(x - 1)^2. Then just replace this in your initial equation and you get that:
x(x - 1)^2 = -5(x - 1)^2.
Then just "shift" -5(x - 1)^2 to the left side of the equation and you get that:
x(x - 1)^2 + 5(x - 1)^2 = 0.
As you can see, 5's sign is now "+", since we've "sent" it to the dark side.... Kidding, the left side :). You will also notice that you can factor (x - 1)^2, since both x and 5 are multiplied by it. This in turn will give you my final equation:
(x + 5)(x - 1)^2 = 0.
Top
Legendary Member
Posts: 2467
Joined: Thu Aug 28, 2008 6:14 pm
Thanked: 331 times
Followed by:11 members

by cramya » Sat Feb 14, 2009 8:22 pm
DanaJ,
Your explanations seem to be spot on in some of the posts I have seen from u.

Keep up the good work as good explanations go a long way in helping everyone out.

Regards,
Cramya
Top
User avatar
Site Admin
Posts: 2567
Joined: Thu Jan 01, 2009 10:05 am
Thanked: 712 times
Followed by:550 members
GMAT Score:770

by DanaJ » Sat Feb 14, 2009 10:32 pm
Thank you, cramya.
Top
Post new topic Post Reply
• Page 1 of 1