Are you sure, it did not say:
X \(\displaystyle \ne\) -2
Well, like I said: I'm unsure. I haven't done math in over 3 years, so more of the deeper meaning or other nuances (that should be common sense at this point, if I had drilled harder in high school) are lost to me. Whether it should have been that or "+2" would of still equated to me not understanding what it would infer.
Stop disparaging yourself. Also don't worry about a placement test. The last thing in the world you want to do is to over-achieve on a placement test and get assigned to a course for which you are not prepared. Being in such a course is a miserable experience that will probably result in a poor grade and perhaps even failure. It's better to take a course that is mostly review and build your confidence than to struggle in panic to get a lousy grade.
I am concerned that you may have copied the problem incorrectly or that the problem is printed incorrectly. I am going to cover some ground that mon ami denis has already covered. In his first post, you notice that he divided both sides of the equation by (x + 2) at one point. That is an entirely legitimate operation unless x = - 2 because then you are dividing by 0, which is not permitted. So the question would make sense if the condition imposed was
\(\displaystyle x \ne -\ 2.\)
Now let's follow the solution
\(\displaystyle (x + 2)^2(x - 1) + 3(x + 2)(x - 1) = (x + 2)P\ and\ x \ne - 2 \implies \dfrac{(x + 2)^2(x - 1) + 3(x + 2)(x - 1)}{x + 2} = \dfrac{(x + 2)P}{x + 2} \implies\)
\(\displaystyle (x + 2)(x - 1) + 3(x - 1) = P.\) This answer, WHICH YOU GOT, is correct as far as it goes, but you then simplified it incorrectly.
\(\displaystyle (x + 2)(x - 1) + 3(x - 1) = P \implies (x - 1)\{(x + 2) + 3\} = P \implies\)
\(\displaystyle P = (x - 1)(x + 5).\) And that is one of the answers given. And even that can be simplified
\(\displaystyle P = (x - 1)(x + 5) \implies P = x^2 + 4x - 5.\)
EDIT: I see Subhotosh Khan beat me to it in wondering if the condition should be \(\displaystyle x = - 2.\) He's too quick for me.
Hmm, about that, this placement is to allow me to take a course which is required for a major I'm interested in (the course I imagine is mandatory), and I didn't know universities placed you in higher courses if you did better on placement exams?
I was also curious as to an alternative solution, since (x - 1)\{(x + 2) + 3\} = P looks quite foreign to me, but after some outside help on the matter, while I was waiting for an answer here, I was told that this was basically a function of distributive property, since ab + cb = (a + c)*b (which I was completely unaware of). My alternative method, which I believe is also adequate; just a bit more work:
(x + 2)(x - 1) + 3(x - 1) = P
I would simply divide out the (x - 1) to the outside, just as I removed the (x + 2) earlier, which would leave me with (x + 2) + 3 = P/(x - 1), then I would re-introduce it back into the equation by multiplying to get the P by itself again: (x - 1)(x + 2) + 3 = P, and finally add the 2 + 3 to get (x - 1)(x + 5). Would that be an alternative/valid way of doing things?
Notice (x - 1) in each term on left; factor it out:
P = (x - 1)[(x + 2) + 3] ; simplify:
P = (x - 1)(x + 2 + 3)
P = (x - 1)(x + 5)
Don't commit suicide...you're doing fine...
I never considered rearranging the result into something more manageable, I was given a tip by some friends that resulted in that solution, which they told me was just rearranging things. I spent about an hour+ talking with them on how that worked, and it seems I'm still missing some groundwork in my basic understanding, as I (once again) never considered rearranging them based on the distributive property, since -
(x - 1) =
a
(x + 2) =
b
3 =
cThis would be translated into - (a)(b) + c(b) = P. And the (a + c)*b portion would equate to - P = (x - 1)[(x + 2) + 3]
Of course, I still feel a tad shook up when that came out to be the case (assuming it is the case), because I couldn't wrap my head around where this came from, so that's when I wondered if I could just factor out the (x - 1) just as I did with the (x + 2)P, and was told that such a method still lead to the same result; it just added some extra work into the mix, but I understood it far better that way, than this way.
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Thanks once again for all the helpful responses! I was also wondering what type of equation that sample problem was, since it introduced some elements that I'm still very unfamiliar with (or am familiar with, but no math for 3 years = lost somewhere in my cranium);
1. (x + 2)^2 = (x + 2)(x + 2), which I shortly learned after posting the original thread was taught in exponents (which I will be covering today, along with Roots).
2. Dividing out both sides to get P by itself. Although it sounds obvious, I had only gone through basic algebraic operations to refresh myself, and polynomials. The reason why I attempted this specific problem was because it "looked" similar to what I was going over, so I decided to try it out. When Denis originally helped me along, it also never occurred to me that the first thing to do was isolate the P, although, looking back on it now, that was the OBVIOUS thing to do, but seeing the x cannot = 2 had me preoccupied, (when in reality, it really shouldn't have, since it had no bearing on the process).
3. Distributive property - Rearranging the simplified solution to simplify it further. That was the biggest hurdle for me. Even though I had people here and elsewhere assisting me with the next step of (x + 2)(x - 1) + 3(x - 1) = P, my mind couldn't understand "how" it reached that stage, and I still have some doubts about it, which is why (if it's correct) I prefer that alternative method, since I understand what's happening, rather than feel like it was some magic trick (which it clearly isn't, but it's synonymous with how I feel it was done).
4. "x
≠ 2 is a
typographical error, on that APTP page. It is supposed to read x
≠ -2." (Quoted from Quaid). I believe in the end, that is the most puzzling thing to me. If it was an error, then how would the correct number factor into the equation?
Ah, but that's not what you wrote. :wink:
x ≠ 2 is a typographical error, on that APTP page. It is supposed to read x ≠ -2.
When x equals -2, the given equation becomes 0 = 0P
In other words, P can be anything, when x is -2
^ I don't entirely understand this statement. If x = -2, then the equation becomes 0? I learned in Absolute Values that x could never be a negative or less than a negative, does that rule also apply here, and is it the same principles governing that rule, or is it applied by another principle?
Afterword -
I'll be looking into Polynomial Division at the start of today (when I wake up), and continue onto Exponent/Roots later in the day. Thank you again, everyone for all the help, and I hope I won't have to return and bother you all again, but if I do, I ask that you please bear with me! I'm sure you don't believe that I'm as inept as I say, but lacking math for 3 years, then having to take an advanced placement exam for a university all of a sudden was jarring to me, and it still is, as even this problem had concepts that I don't consciously "know", but sub-consciously "do know", I just haven't gotten to Exponents yet (until today), so I think I'm at least slightly entitled to say that I'm a bit slow on the "up-take" as it were.
