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It's a factoring question.
oh. yeah -(24a^2+6a-9) take out the common factor 3 -3(8a^2+2a-3) so this is a "hard factoring problem"
try... 2 and 4 for a, and -3 and 1 for the constant.
I'm really confused by this. Would appreciate greatly any help. :-)
I'm a bit confused about what you did, hedgepig. Perhaps I need to review your text for a bit. If you could, please do stay. :-)
Okay, so what is the way to figure out the order?
first, you want to factor out the largest constant.
What comes first? A, AB, or B? This is just in general.
Is a squared before a?
9-6a-24a^2 factor out the largest number you can first.
Yes, but I guess I just need to know, in general, the rule for how you order things. Because, I have more problems, of course.
So, it's just, if you wouldn't mind, I'd like to know how to properly order this trinomials.
It's so important.
there are multiple ways to process this, but, we can do that too. Order from highest degree to lowest.
But I don't what's highest. Is a^2 higher or lower than a?
is it? what is the degree of a^2?
a^2 is a x a, but I don't know whether it would come before or after a.
If I were to guess, I would say after.
I would say a exponents, followed by B. I would guess ab would come before B, too.
from what I gather you're missing knowledge from polynomials and terminology. Do you what the degree of a term is?
Unfortunately, I am not very knowledgeable on the terminology.
Ok. You NEED to know this stuff. that being said, it shouldn't be too hard. let's do some examples: 6x^3 23x^5 x^3 8x 9 -5x^2 identify the degree of each term.
I would guess the degree is the exponent.
So, degrees are synonymous with exponents?
what we are referring to is, as far as I can tell, a specific realm - the realm of polynomials.
to answer your question, yes, degree is synonymous with exponents in the context of polynomials, but I am not sure I would say x^(-pi) would have degree -pi.
In the poly and trinomial universe, degrees and exponents are the same thing.
That's the universe my head is in right now. :P
that's not a problem in this context, but I would really advise you think broader than this if you want success mathematically.
Now, I really do need to know how to order these trinomials, because they're giving them to me out of order. If I can understand how the order works, it should be easy.
i already gave you the answer to this - you order them from highest degree to lowest.
The problem is, I genuinely do not know what the highest degree is. I don't.
note that for expression x^4 y^2 z^7 the degree is 4+2+7=13
Yes. That is, again, why i told you to identify the degree of each term.
6x^3 23x^5 x^3 8x 9 -5x^2
So, 24a^2 is first
of what degree? and you forgot a minus sign.
The other two don't have a degree, though.
Yes, my bad.
I don't understand why you keep ignoring me. Identify the degree of each term 6x^3 23x^5 x^3 8x 9 -5x^2
The degree is the 2.
I'm sorry, I'm not trying to ignore you.
The degree isn't the 2 in that case. The degree IS 2.
1. 3, 2. 5, 3.3, 4. IDK, 5. IDK 6. 2
5 is 1, and 6 is 0.
it might be hard (I know I had a bit of trouble with exponents in the beginning) but you really need to know how to identify and visualize what's going on here.
9-6a-24a^2 Is -24a^2-6a+9
for term ax^n for constant a and n, the coefficient is equal to a and the degree is equal to n for term bx^n y^m the coefficient is b and the degree is equal to n+m
yes... but, what I don't understand is how you are able to factor without knowing how to order polynomial terms
Perhaps that was the only reason I was having confusion. I would like to work on the problem a bit, see if I get it right, and you can confirm.
Well, I guess we'll see if I know how to factor. I think I do, though. :)
the answer is aways above if you wish to check it.
Yes, but I do want to do it myself.
my suggestion to you, if you are so dedicated and inclined, and if you wish to have future success in mathematics, is to review prealgebra
there are plenty of resources available online for free to do so...
Anyways, if I did this correctly, I have to think of two factors that add to -6 and when multiplied, are -216. Is this what I am supposed to do?
where are you getting -216
24 x 9
-24 x 9
Okay, so... am I to square the first and last number?
actually, let me think, I have never learned how to factor that way.
Well, 24 isn't a perfect square. So, honestly, I have no idea what to do from here.
nope, that doesn't look right at all.
I was walking you through the problem step by step, but I must've gone too quickly.
Do you understand the purpose of factoring trinomials, and what the inverse operation of that is?
My understanding is that the purpose of factoring is to simplify.
The inverse operation, I would assume how we're all familiar with foil, well the inverse operation would be the system to get a trinomial in into the state of foil. How it's implemented, though, I'm not sure.
well why don't we look at it. let's foil (ax+b)(cx+d) what do you get?
Give me a second.
I got acx^2+axd+bcx+bd
ok. that's equal to (ac)x^2+(ad+bc)x+(bd) right?
remember our expression -24a^2 -6a+9 ? here is what factoring really is. ac=-24 -6=ad+bc 9=bd Solve the system for a, b, c, and d
I don't understand the question.
Note that, once we have the solution, we know that the factored form of the expression is simply (ax+b)(cx+d)
You want me to tell you which each letter signifies individually? I don't know how to do that.
no. I just gave you a system of equations.
ac=-24 -6=ad+bc 9=bd
Yeah, I don't understand what that is asking.
find values of a, b, c, and d such that each of the three equations holds true
Will do. :-)
Give me a minute, please. Thank you.
this is how I got the system by the way: (ac)x^2+(ad+bc)x+(bd) -24a^2 -6a+9 the variable a is redudant but for the second equation take x=a to get (ac)x^2+(ad+bc)x+(bd) -24x^2 -6x+9 well, (ac)x^2+(ad+bc)x+(bd) (-24)x^2+(-6)x+(9) thus, (ac)=(-24); (ad+bc)=(-6);(bd)=(9)
by the way, here is a hint: you should not try to solve the system of equations using algebra. Use trial and error.
Okay. :-) I knew that.
A = -6
C = 4
B = 3
D = 3
and what will our factored expression be?
I don't know.
what did I tell you to foil?
Sorry, I do get stressed when I work on math. My brain usually isn't at its peak at this point.
you are doing just fine - I am not approaching this problem in the standard way that most people learn how to factor... I choose not to do so because you will learn the standard way from everyone else...
You wanted me to foil (ax+b)(cx+d)
ok... so simply resubstitute your values for a, b, c, and d, no?
What do you get?
Give me a moment.
Okay, I don't understand. I didn't figure out the meaning for an X.
I do appreciate you trying to be innovative in your explanation, but perhaps the standardized explanation is so for good reason?
mmm this is a key point that most people struggle with algebra. We first introduce letters that represent numbers in algebra. Well, we often tell people to evaluate expressions by assigning them values for particular letters. Here's the thing though - there are basically two kinds of symbols: symbols that represent CONSTANT numbers and symbols that represent UNKNOWN numbers. You might wonder, why is it that we need symbols for CONSTANT numbers? Consider the polynomial of degree two ax^2+bx+c There are infinitely many polynomials of this form. here are a few. x^2+3x-3 10x^2-12x+5 but what about x? we don't actually have any fixed value for x, so we leave it as is, despite knowing what a, b, and c are. In this case, i have given you an expression with a, b, c, d, and x and we have found values for a, b, c, d... but not x... why is it that we do not know what to do for x? Simple. Let's look at the original problem. 9-6a-24a^2 note that halfway through I turned the a's to x's. No matter. From the start, we defined a (x right now) as a symbol that varied. There is no way to find it.
The standardized explanation is why many of my peers struggled deeply with calculus - not because they could not follow the rules of calculus, but because they could not CREATE expressions from understanding algebraic motivations.
I am not saying not to look at the standardized explanation - I have a pretty good site I will refer you to after this, but we are almost at the end of this exercise.
Yeah, by all means continue. You know what you're doing. :-)
-24^2-6a+9 I just don't know how to factor it.
for (ax+b)(cx+d) the substitution a = -6 c = 4 b = 3 d = 3 yields? do not worry about foiling the result
Okay. I will do that.
You basically just factored the expression. Now, there is one last step we must perform to call out expression factored. We can actually factor 3 out from the first parenthetical
So it would be (-2x+3)(2x+3)-3
yes.. but when you have a factored expression, you actually put any constants to the very front. so your -3 belongs in the front of everything.
and I think you meant, for the second parenthetical, 4x+3
Okay, so how do I do with a trinomial?
what do you mean?
The original problem
I still honestly don't know what to do.
I just became you're first fan, by the way.
hahaha I am on an alt, but, yes, first fan on this account :)
(-2x+3)(2x+3)-3 move the negative three to the other side
erm, 4x+3 sorry
sorry we are really close to the answer and I don't want to steal it from you by typing it ahaha
the second parenthetical should be 4x+3... you miscopied it before and I did the same
Did I do it correctly?
yeah... you just went about my roundabout method of factoring the expression (and you got it right! though those x's are supposed to be a's :P http://www.wolframalpha.com/input/?i=factor+9-6a-24a%5E2+ ). if you wish, you can remultiply to ensure that it is correct. but, as I promised, here are some links to help you with factoring this is the homepage of the site I learned algebra from http://www.purplemath.com/modules/index.htm http://www.purplemath.com/modules/factquad.htm here is the famous khan academy https://www.khanacademy.org/math/algebra/introduction-to-polynomials-and-factorization/factoring-polynomials-2-quadratic-forms/v/factoring-polynomials-1
pertaining DIRECTLY to what we just did is http://www.purplemath.com/modules/factquad2.htm
Okay, but I still don't know how to do the original problem, it's a trinomial.
all of those links deal with trinomials. quadratic refers to the fact that they have maximum degree 2.
You know, I sometimes do need a little bit of a different approach.
Could you tell me what is the step you would take in solving the original problem now that we know the proper order?
note that this is also called factoring by grouping
Okay, I saw the next step. I didn't see the answers. I'll like to see if I got it right. :-)
So, let's see if I get this! :-)
mmhmm - I think this site has the lesson you are looking for fyi http://www.purplemath.com/modules/factquad2.htm
Working on it. Hope you stay. :-) Thank you so much for your time.
Okay, did -3 HAVE to be factored out? My natural proclivity was to factor 3.
haha no problem -thanks for bearing with me even when it hasn't been apparent where I've been taking the discussion. unfortunately, it's around 2:50AM and I have a chemistry class tomorrow at 9AM, so I should probably head off to bed. As for -3, it is standard to make the leading coefficient (the number in front of the a^2 in this case) to be positive, because otherwise all the other signs have to be switched while you try to factor.
the point of using -3 was to make it so the number in front of a^2 was positive
Okay. Are you in college? What year are you?
I'm a freshman in college.
Ah, okay. Yeah, I better let you sleep. It is important.
best of luck to you - feel free to respond to those post if you have anymore questions, though I would actually suggest you open a new question with the same problem and have some other persons look at it - they will give a much more straightfoward way of explaining how to factor.
What's your major?
undeclared engineering - are you in college?
Nope. Junior in high school.This is Algebra 2
Does my work seem college level? :p
No, but you would be surprised as to the number of students who take "college algebra" which is merely junior high school algebra in college. Your work does not seem college level, but your vocabulary does seem so.
I'm a bit confused as to what to do with +6 and -4
Could you illustrate that?
hmm one moment
I'm sorry, I hate to take up your sleep, but hopefully, once I'm certain how to do one question, I'll know how to do them all.
so, you have the expression (8a^2+2a-3) well, let's look at (ax+b)(cx+d)=(ac)x+(cb+ad)x+(bd) note that this is saying we search for two numbers that have a product of 8. (ac=8) we search for two numbers that have a product of -3 (bd=-3) we seek to make ad+bc equal to 2. this is where the 6 and -4 come in. 6-4=2, which means ad=6 and bc=-4, or ad=-4 and bc=6.
why did we pick 6 and -4? well, it was a natural choice because if we look at the factors of 8, we have 2 and 4 and 1 and 8. if we look at the factors of -3, we have 1 and -3 or -1 and 3
Yes, I know how to get the factors, but I don't know what to do with them.
well, there are a lot of choices for the factors of 8, so we will worry about the factors of three instead.
thus we wind up with something like (?a+3)(?a-1)
well, we will wind up with 3?a-?a such that we get 2a when we pick two question marks. note the question marks have to multiply to 8. because ?a*?a=8a^2
My guess, would be (8a^2+6) (4-3) (1) 4-3 is 1, which we can cross off, so -3(8a^2+6)
hold on... the answer -3(8a^2+6) is not what we originally looked at. if you simplify that, you get somethign different fromt he original expression.
I don't know, going to need your explanation skills 'cause I'm lost.
Says you've been typing for a while.
let's start over. we just factored out that -3 -3(8a^2+2a-3) so let's worry instead about factoring 8a^2+2a-3. Well, all this is asking you to do is to rewrite 8a^2+2a-3 in the form (ra+t)(qa+s) well, factoring is kinda one of those things wher eyou guess and check. we can foil out our expression to visualize it better though. we'll get (rq)a^2+(tq+rs)a+ts compare it with 8a^2+2a-3 in note that 8 HAS to equal rq and -3 HAS to equal ts and 2 HAS to equal tq+rs. Why? because our expression is in the most simplified form possible. we know that all the like terms have been combined, so we can match the coefficients safely. it then boils down to solving the system. to do so, let's compare the pairs rq and ts and see which one has fewer potential factors. the factors of 8 are 1,2,4,8 (or -1,-2,-4,-8) while the factors of 3 are (-1,1,-3,3) so let's start with picking arbitrary factors of three for our values ts. so, let's say t=3, and s=-1 then, we have (ra+t)(qa+s) -> (ra+3)(qa-1)=8a^2+2a-3 well we know they already multiply to -3 because we chose them to. however, we're gonna have to make additional constraints to this problem if we want to get anywhere. let's pick values for r and q. remember rq=8, so let's just do r=2 and q=4 then (2a+3)(4a-1)... well we are really just trialing and erroring for the value tq+rs. in this case we get tq=12 and rs=-2 tq+rs=10. We're supposed to get 2. so, let's think... how can we get 2? well, it just so happens that if tq=6 and rs=-4, we'll get 2. That intuition comes with expreience, but that is the origin of the 6 and -4. knowing that, it follows that because ts=-3, t=3, and q=2. whereas, s=1, and r=-4 that is the solution in this case, which you deducted earlier actually.
Okay, I know how to figure out the factors. That's easy to me.
then what isn't?
6 and -4. I struggle with what to do with the factors, though.
That is my struggle.
I'm fine with the guess work.
8a^2+2a-3 and we have 6 and -3. I don't know what to do from here.
just resubstitute. again, with the expression (ax+b)(cx+d) you will get acx^2+(ad+bc)x+bd...
your 6 and your -4 are, respecitvely, your ad and your bc in this instance.
So (8a^2+6) and (-4-3). Is that right?
... no - foil the result and you will quickly learn that that does not yield what you need it to I am not sure i can do a good job explaining this... I would advise you check out some videos on khan academy. I am basically out of time but not done explaning, so I will give you this video which uses another method (factoring by grouping) and hope it will help. It's 5 minutes... https://www.khanacademy.org/math/algebra/introduction-to-polynomials-and-factorization/factoring-polynomials-2-quadratic-forms/v/factoring-trinomials-by-grouping-6 if you have time, I'll be online very shortly tomorrow morning.
(in around 4 hours)
Funny you say that, as I am studying French.
If I could have I would've asked you to show your work that got you to that result, but unfornuately I really am out of time - no point in me showing up to chem class if I can't even think
Fair enough. Thank you so much for staying as long as you did; I'm sure it was frustrating dealing with me.
I "studied" french - all i remember are the very, very basics now, unfortunately. There was no frustration dealing with you by the way - it is incomparable how often I get frustrated with entitled kids who come on this site demanding answers with neither appreciation nor patience. You have a good degree of both, and you are one of the rare cases where the teacher leaves before the student. I will admit that I am somewhat frustrated though - not at you for not understanding the concept, but at myself for not being able to articulate that which i supposedly find so straightfoward and easy to do. Farewell!