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#1
08-08-2010, 04:50 PM
 friedo Charter Member Join Date: May 2000 Location: Brooklyn Posts: 19,809
Embarrassingly simple algebra help required

I have to take a placement test in a few weeks and I've been reviewing math. Apparently I've forgotten everything after counting.

How do you factor this polynomial: (x+3)2 - 4(x+3)

The answer in the back of the book is (x-1)(x+3). I think this is supposed to be a "difference of two squares" one but I can't figure out how to express -4(x+3) as a square. All I've been able to do so far is move various (x+3)s around, FOIL out (x+3)2, distribute -4(x+3), and none of that has gotten me anywhere.
#2
08-08-2010, 05:02 PM
 tim-n-va Guest Join Date: May 2007
(x+3)^2- 4(x+3)
(x+3)(x+3) - 4(x+3)
(x+3)((x+3)-4)
(x+3)(x-1)
#3
08-08-2010, 05:07 PM
 friedo Charter Member Join Date: May 2000 Location: Brooklyn Posts: 19,809
I still don't understand. Where does ((x+3)-4) come from and how does this turn into (x-1)?
#4
08-08-2010, 05:07 PM
 Johnny L.A. Charter Member Join Date: Jan 2000 Location: NoWA Posts: 46,499
(x2 + 6x + 9 ) – (4x + 12)

x2 + 6x - 4x + 9 - 12

x2 + 2x - 3

(x - 1)(x + 3)

EDIT: Dammit, tim-n-va!

.

Last edited by Johnny L.A.; 08-08-2010 at 05:08 PM.
#5
08-08-2010, 05:07 PM
 Exapno Mapcase Charter Member Join Date: Mar 2002 Location: NY but not NYC Posts: 22,166
Quote:
 Originally Posted by friedo I have to take a placement test in a few weeks and I've been reviewing math. Apparently I've forgotten everything after counting. How do you factor this polynomial: (x+3)2 - 4(x+3) The answer in the back of the book is (x-1)(x+3). I think this is supposed to be a "difference of two squares" one but I can't figure out how to express -4(x+3) as a square. All I've been able to do so far is move various (x+3)s around, FOIL out (x+3)2, distribute -4(x+3), and none of that has gotten me anywhere.
Or this more general route.

First, multiply out the two components.

(x+3)2 - 4(x+3) = x2 + 6x +9 -4x -12

x2 +2x -3

Factor.

(x+3)(x-1)
#6
08-08-2010, 05:10 PM
 John Mace Guest Join Date: Dec 2002
Quote:
 Originally Posted by friedo I still don't understand. Where does ((x+3)-4) come from and how does this turn into (x-1)?
ab + ac = a(b+c)

or

aa + ac = a(a+c)

a = (x+3); c = -4

Last edited by John Mace; 08-08-2010 at 05:12 PM.
#7
08-08-2010, 05:11 PM
 tim-n-va Guest Join Date: May 2007
Quote:
 Originally Posted by friedo I still don't understand. Where does ((x+3)-4) come from and how does this turn into (x-1)?
(x+3) was a common factor in both terms in the previous line. ((x+3)-4) was there to show the step but the inner parenthesis has no effect so it can be re-written as (x+3-4) or (x-1).

The expanding, simplifying and refactoring in the other answers works and gives the same result.
#8
08-08-2010, 05:22 PM
 friedo Charter Member Join Date: May 2000 Location: Brooklyn Posts: 19,809
Quote:
 Originally Posted by Johnny L.A. (x2 + 6x + 9 ) – (4x + 12) x2 + 6x - 4x + 9 - 12 x2 + 2x - 3 (x - 1)(x + 3) EDIT: Dammit, tim-n-va! .
Quote:
 Originally Posted by Exapno Mapcase Or this more general route. First, multiply out the two components. (x+3)2 - 4(x+3) = x2 + 6x +9 -4x -12 Add and subtract like exponents. x2 +2x -3 Factor. (x+3)(x-1)
These two methods make sense, thanks. I'm still missing something and don't quite get tim's method, but I'll work through it.

Regardless, I think I am totally, utterly fucked when it comes to this test.
#9
08-08-2010, 05:28 PM
 John Mace Guest Join Date: Dec 2002
Friedo:

tim's method is the most elegant. You see that the two terms have a common factor: (x+3). So factor that out of the square.

(x+3)^2 = (x+3)(x+3).

Then you use the rule I posted above: ab + ac = a(b+c) where a = (x+3), b= (x+3) and c = -4
#10
08-08-2010, 05:31 PM
 Pasta Charter Member Join Date: Sep 1999 Posts: 1,567
The "multiply it all out and then divine the two factors" method is okay, but it is fairly brute force. The key here is to recognize that the two terms in the expression already have a common factor in them: (x+3).

(x+3)2-4(x+3) = (x+3)(x+3)-4(x+3)

Pull out a factor of (x+3) from both terms to get your first factor handed right to you:

(x+3) [ (x+3) - 4 ]

Now all you need to do is comine the 3 and -4 in the brackets:

(x+3) [ x + 3 - 4 ] = (x+3)(x-1)
#11
08-08-2010, 05:36 PM
 Thudlow Boink Charter Member Join Date: May 2000 Location: Springfield, IL Posts: 16,682
Quote:
 Originally Posted by friedo I'm still missing something and don't quite get tim's method, but I'll work through it.
How would you factor A2 - 4A?

Answer: Factor out the common factor A: A(A-4).

That's exactly what's going on here, except that A is really x+3 in disguise.

Quote:
 Regardless, I think I am totally, utterly fucked when it comes to this test.
If it's a placement test, it should be designed to make sure you don't end up in a class that's too hard or too easy for you. Don't think of it in terms of being "fucked"; think of it in terms of finding out what you know vs. what you still need to learn.
#12
08-08-2010, 05:43 PM
 Derleth Guest Join Date: Apr 2000
Quote:
 Originally Posted by friedo Regardless, I think I am totally, utterly fucked when it comes to this test.
The point of a placement test is to get your fucking done all in one spot and quickly, as opposed to drug out over the course of an entire semester you really weren't prepared for. Don't worry too much about the test.
#13
08-08-2010, 05:45 PM
 Johnny L.A. Charter Member Join Date: Jan 2000 Location: NoWA Posts: 46,499
Quote:
 Originally Posted by John Mace tim's method is the most elegant.
Quote:
 Originally Posted by Pasta The "multiply it all out and then divine the two factors" method is okay, but it is fairly brute force.
Quote:
 Originally Posted by friedo Regardless, I think I am totally, utterly fucked when it comes to this test.
Elegance is nice, but brute force done step-by-step leaves less room for error, I think. (By 'error' I mean misremembering or misapplying rules. In more complex equations, there's always the chance of a boneheaded error like writing down the wrong sign or something.)
Quote:
 Originally Posted by Thudlow Boink If it's a placement test, it should be designed to make sure you don't end up in a class that's too hard or too easy for you. Don't think of it in terms of being "fucked"; think of it in terms of finding out what you know vs. what you still need to learn.
Agree. It's better to go back over the stuff you already know to make sure you have the fundamentals, than to charge ahead and struggle with things you're forgotten.
#14
08-08-2010, 05:51 PM
 aceplace57 Guest Join Date: Oct 2009
It pays to keep an old algebra book from college. This stuff comes back quickly with some review. I've had to scrape the rust off my memory a few times since college. It really helps to use the book that you originally studied from.

Last edited by aceplace57; 08-08-2010 at 05:52 PM.
#15
08-08-2010, 05:53 PM
 John Mace Guest Join Date: Dec 2002
Quote:
Originally Posted by Johnny L.A.
Quote:
 Originally Posted by John Mace tim's method is the most elegant.

Elegance is nice, but brute force done step-by-step leaves less room for error, I think. (By 'error' I mean misremembering or misapplying rules. In more complex equations, there's always the chance of a boneheaded error like writing down the wrong sign or something.)
Quote:
 Originally Posted by Thudlow Boink If it's a placement test, it should be designed to make sure you don't end up in a class that's too hard or too easy for you. Don't think of it in terms of being "fucked"; think of it in terms of finding out what you know vs. what you still need to learn.
Agree. It's better to go back over the stuff you already know to make sure you have the fundamentals, than to charge ahead and struggle with things you're forgotten.
In this case, the elegant method is the simpler method, too. The brute force method takes longer and is more complicated-- open to more errors.
#16
08-08-2010, 05:57 PM
 Pasta Charter Member Join Date: Sep 1999 Posts: 1,567
Quote:
 Originally Posted by Johnny L.A. Elegance is nice, but brute force done step-by-step leaves less room for error, I think.
I agree that getting the answer right is usually all that matters. In a test-taking environment, though, it is not always the case that "minimize errors" is optimal. Generally, speed it Priority 1. Problems like this one are not looking to test whether you can factor the polynomial but whether you can factor it quickly by recognizing the common factor straight away, thus demonstrating an intuition about the factoring concept. A whole test's worth of brute force approaches will likely result in a lower score.

Not trying to scare the OP -- just saying that that's how I view this particular problem on a placement test.
#17
08-08-2010, 07:33 PM
 Indistinguishable Guest Join Date: Apr 2007
Quote:
 Originally Posted by Johnny L.A. Elegance is nice, but brute force done step-by-step leaves less room for error, I think.
Understanding is best. If you cannot see at a glance how to factor out the common (x + 3), it doesn't matter if you can still manage to work your way to solving the problem by "brute-force": at this point, a serious hole in understanding has already been revealed which needs filling in before advancing further. Solving the problem is not the main issue; no one needs the problem solved and you're not going to get a prize for doing so. The whole point of the exercise is to accurately assess the contours of your understanding so as to know how to appropriately educate you further. It's not meant to be "What are you able to do?"; it's meant to be "What do you understand?". It just happens to be easier to assess the former to use as a proxy for the latter.

(Not that this should mean anything in terms of how you take placement tests. It should just mean something in terms of how you react, in your own education, to, say, a similar situation arising as you attempt homework)

Last edited by Indistinguishable; 08-08-2010 at 07:37 PM.
#18
08-08-2010, 08:22 PM
 needscoffee Guest Join Date: Oct 2009
freido, here is a great, free site for math review and help (not to mention chemistry, physics, biology and more): KhanAcademy.org . There are over 1600 lessons/demonstrations there now, from arithmetic through advanced college.

Last edited by needscoffee; 08-08-2010 at 08:22 PM.
#19
08-08-2010, 08:35 PM
 friedo Charter Member Join Date: May 2000 Location: Brooklyn Posts: 19,809
Quote:
 Originally Posted by Johnny L.A. Agree. It's better to go back over the stuff you already know to make sure you have the fundamentals, than to charge ahead and struggle with things you're forgotten.
I took AP Calculus in high school. I work full time now. I don't have the time to spend a year in remedial math courses because my brain has become swiss cheese.

Anyway, thanks for the help. I've been spending the afternoon doing more factoring exercises from the book I have and it's starting to come back to me more quickly.

I'm not worried about taking calculus again (assuming I can get by the %*&#ing test.) Calculus is easy. It's algebra that wants to rape my brain. Every mistake I ever made in high school calculus was an algebra mistake.
#20
08-08-2010, 08:49 PM
 pulykamell Charter Member Join Date: May 2000 Location: SW Side, Chicago Posts: 27,702
This may be bleeding obvious, but in high school it didn't seem to be for a lot of people: if you have the time, make sure to check your answer by plugging in an arbitrary number for X (I like to pick 2 or 3--whatever number doesn't introduce weirdness like dividing by zero and the such) and check to see if your factored answer matches the original equation. It won't necessarily tell you if you're absolutely right, but a mismatch in answers will indicate you're wrong.
#21
08-08-2010, 08:52 PM
 Indistinguishable Guest Join Date: Apr 2007
Quote:
 Originally Posted by friedo I'm not worried about taking calculus again (assuming I can get by the %*&#ing test.) Calculus is easy. It's algebra that wants to rape my brain. Every mistake I ever made in high school calculus was an algebra mistake.
Just out of curiosity, what do you think it is that makes algebra so much more difficult for you than calculus?
#22
08-09-2010, 01:04 AM
 friedo Charter Member Join Date: May 2000 Location: Brooklyn Posts: 19,809
Quote:
 Originally Posted by pulykamell This may be bleeding obvious, but in high school it didn't seem to be for a lot of people: if you have the time, make sure to check your answer by plugging in an arbitrary number for X (I like to pick 2 or 3--whatever number doesn't introduce weirdness like dividing by zero and the such) and check to see if your factored answer matches the original equation. It won't necessarily tell you if you're absolutely right, but a mismatch in answers will indicate you're wrong.
Yep, I've been doing that. A lot of the exercises end up with a pair of binomials which is easy to check by FOILing them, so I've got that down. I also use a small integer to check things out when I'm not 100% sure that a given manipulation is valid. The rules are beginning to come back to me. I should have started reviewing earlier, but I was in class for the first half of the summer on top of my job. And then I went to Las Vegas for a few days........

Quote:
 Originally Posted by Indistinguishable Just out of curiosity, what do you think it is that makes algebra so much more difficult for you than calculus?
I think it's just because algebra is a huge subject. There are so many different things and sets of rules that you have to remember and practice. Calculus (the basic stuff anyway) is just learning derivatives and integrals and then spending the rest of the time learning how to apply those two things to different kinds of problems. Most high school algebra programs are actually two years long and they don't cover nearly everything that you'll need to advance.

Honestly, most of algebra isn't that hard (I've been feeling sorry for myself again). It's just that there are a few items that I never mastered (factoring polynomials are among them.) And the book I'm working from is comprehensive but frustratingly terse.
#23
08-09-2010, 01:22 AM
 Indistinguishable Guest Join Date: Apr 2007
Ah. Yes, any subject is daunting when it seems like an ad hoc collection of disparate rules.

For what it's worth, I think of algebra as also just a few basic elements that you gain experience applying in many situations: the rules of addition, multiplication, and occasionally exponentiation of variables. There may be a few other things covered besides these, but the basic rules are quite simple; I could give ten or so rules that would suffice to work out 90% of all algebra problems (things like "The order of addition doesn't matter", "a * (b + c) = a * b + a * c", and so on). The only difficulty is in developing the intuitions to know what kind of manipulations to try in working out a particular problem, and that just comes with practice.

Last edited by Indistinguishable; 08-09-2010 at 01:24 AM.
#24
08-09-2010, 05:58 AM
 congodwarf Guest Join Date: Aug 2003
freido, A few months ago when I was preparing to take my placement test, someone directed me here -Khan Academy. My practice test placed me in remedial math.

I spent about 2 weeks reviewing everything from 1+1=2 through college algebra. After those 2 weeks, I placed into college level algebra.

I haven't taken a math class since 1997 and the highest math I completed was Algebra 2/Trig.

I'll be taking both my required math classes next semester (statistics and something else, don't remember what) and I don't have to waste any money on refresher courses.
#25
08-09-2010, 08:16 AM
 tim-n-va Guest Join Date: May 2007
Quote:
Originally Posted by Pasta
Quote:
 Originally Posted by Johnny L.A. Elegance is nice, but brute force done step-by-step leaves less room for error, I think.
I agree that getting the answer right is usually all that matters. In a test-taking environment, though, it is not always the case that "minimize errors" is optimal. Generally, speed it Priority 1. Problems like this one are not looking to test whether you can factor the polynomial but whether you can factor it quickly by recognizing the common factor straight away, thus demonstrating an intuition about the factoring concept. A whole test's worth of brute force approaches will likely result in a lower score.

Not trying to scare the OP -- just saying that that's how I view this particular problem on a placement test.
I took the SATs 35 years ago so my memory for the details isn't perfect. However, this point was exactly the impression I had from the math portion. In many of the problems there was the obvious brute-force way to get the solution but if you really understood you'd see a short cut. On a timed test that was essential.
#26
08-09-2010, 12:10 PM
 friedo Charter Member Join Date: May 2000 Location: Brooklyn Posts: 19,809
Quote:
 Originally Posted by congodwarf freido, A few months ago when I was preparing to take my placement test, someone directed me here -Khan Academy. My practice test placed me in remedial math.
Wow, cool site. I'll be sure to check it out.
#27
08-09-2010, 12:33 PM
 needscoffee Guest Join Date: Oct 2009
Quote:
Originally Posted by friedo
Quote:
 Originally Posted by congodwarf freido, A few months ago when I was preparing to take my placement test, someone directed me here -Khan Academy. My practice test placed me in remedial math.
Wow, cool site. I'll be sure to check it out.
What am I, chopped liver? Did no one see post #18?
#28
08-09-2010, 02:13 PM
 congodwarf Guest Join Date: Aug 2003
In my defense, it was really really late when I posted. I skimmed the thread before posting but it was just too late to read all that math.

Awesome site isn't it? I've passed the recommendation on to so many people, including the program coordinator for the liberal parts program at my school (you have to place into college algebra to get into the program).
#29
08-09-2010, 03:32 PM
 needscoffee Guest Join Date: Oct 2009
Quote:
 Originally Posted by congodwarf In my defense, it was really really late when I posted. I skimmed the thread before posting but it was just too late to read all that math. Awesome site isn't it? I've passed the recommendation on to so many people, including the program coordinator for the liberal parts program at my school (you have to place into college algebra to get into the program).
I was referring more to the OP.

That site is astonishing. That guy does more before breakfast than I do all year.
#30
08-25-2010, 04:49 PM
 friedo Charter Member Join Date: May 2000 Location: Brooklyn Posts: 19,809
Just wanted to thank everyone for the answers and pointers provided. After spending most of my spare nights and weekends reviewing all the algebra and trig that I have forgotten, I totally kicked the test's ass and now have three semesters of calculus to look forward to.

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