# 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)[sup]2[/sup] - 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)[sup]2[/sup], distribute -4(x+3), and none of that has gotten me anywhere.

(x+3)^2- 4(x+3)
(x+3)(x+3) - 4(x+3)
(x+3)((x+3)-4)
(x+3)(x-1)

I still don’t understand. Where does ((x+3)-4) come from and how does this turn into (x-1)?

(x[sup]2[/sup] + 6x + 9 ) – (4x + 12)

x[sup]2[/sup] + 6x - 4x + 9 - 12

x[sup]2[/sup] + 2x - 3

(x - 1)(x + 3)

EDIT: Dammit, tim-n-va!
.

Or this more general route.

First, multiply out the two components.

(x+3)[sup]2[/sup] - 4(x+3) = x[sup]2[/sup] + 6x +9 -4x -12

x[sup]2[/sup] +2x -3

Factor.

(x+3)(x-1)

ab + ac = a(b+c)

or

aa + ac = a(a+c)

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

(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.

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.

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

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)[sup]2[/sup]-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)

How would you factor A[sup]2[/sup] - 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.

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.

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.

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.)

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.

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.

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.

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.

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)

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.

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.

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.