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Tutor question #2
Factor: x^4 7x^2 +1
(conditions: pen, paper, no calculator, done in 2 minutes)
**
 one year ago
 one year ago
Tutor question #2 Factor: x^4 7x^2 +1 (conditions: pen, paper, no calculator, done in 2 minutes) **
 one year ago
 one year ago

This Question is Closed

maheshmeghwal9Best ResponseYou've already chosen the best response.0
May be we can take x^2 as "t" & then factor
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
I mean\[(t^27t+1+5)5\]
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
If my brain still works, Then it becomes (t6)(t1)5 and then ..?
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
I should learn this :/
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
ya it is not going to be factor like this:/ May be @satellite can help us:)
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
so we made only this much\[(x^26)(x^21)5.\] now what should we do of {5}?
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
I don't think we should do it in this way...
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
ya u r right @Callisto But I m thinking more on this.
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
you need to use synthetic division
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
@bronzegoddess Can you further explain it?
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
x^26 @ParthKohli
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
@ParthKohli It's not like that.... and you made mistakes there too...
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
I deleted it before you noticed :P
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
I noticed before you deleted :P
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
This degree 4 is eating my head >.<
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
I m feeling giddy too
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
@maheshmeghwal9 can you call Yogesh please?
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
but he is sitting aside of me
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
we 3 are all here:)
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
okay, lets make a guess that (x2) is a factor then we'll use synthetic division to actually se eif it one, its a trial and error method though: dw:1339672516505:dw so according to this division(x2) isnt a factor because it gives us a remainder 11.
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
But my wolfy says (x^2  3x + 1)(x^2 + 3x + 1)
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
@bronzegoddess According to the word 'factor', I don't expect there is a remainder.
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
ya @Callisto is right: D
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
@ParthKohli I said no wolf. I got that answer from wolf too. But that doesn't help you solve it because you cannot use calculator. Just pen and paper
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
Hmm but I was telling bronzegoddess that this can be factored. I wanna understand too
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
LANA THE PROFESSOR :D
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
Not using synthetic division like that  yes :) Lana~~~~~~~~~~~~
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
@Callisto as I said it is a trail and error, and i was giving you an example, of course you have to start with finding the possible factors using descartes rule of signs!
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
We must want an exact & good method.
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
okay, let me do the whole thing again.
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
@bronzegoddess If the factor is 2 terms like ax+b, I can do it. But in this case, it's not. So I don't know how. If you don't mind, can you teach me how to reach that answer?
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
very good @ajprincess
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
but how did she know about adding 3x^3 & something other. I want to hear from her.
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
ya but if we didn't know the answer, then what to do It is the exact problem:/
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
ya will show that in a few minutes time.
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
first: number of possible roots is 4 because of the degree. second Descartes' Rule: number of +real roots = 1or zero, and real roots= 1 or zero, imaginary=2 or 4 third: what are the possible roots; constant term/leading coefficient so \[\pm1/\pm1\] therefore your 4 possible roots are 1,+1 and 2 imaginary numbers. fourth: synthetic division to see if 1 and +1 are actual roots; dw:1339673466435:dw but since it gives us a remainder then it isn't a factor(x+1) then you do the same with(x1) and if it also give you a remainder then all your factors are imaginary. Thats how I learnt it.
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
Do you know when to use that method?
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
we use trial and error method when we need to factorise expressions with greater power
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
when looking for factors or zeros of polynomials
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
@bronzegoddess please take a look at this: http://en.wikipedia.org/wiki/Factor_theorem We can apply that method when one of the factors is in the form (xk) Now, in this case, no factors are in the form (xk). So, it cannot be applied in this case.
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
Of course you can keep using that method for checking, but you can't get a factor, I think.
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
@Callisto (xk) aka (x1)=0 is x=1 which you'll end up using as the divisor in synthetic division.
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
Okay, to be simple, if you can get a factor using that method, please tell me.
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
my step by step explanation from before doesn't make sense?
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
Sorry, to be honest, it makes no sense to me in this case. I truly understand that method and have been using to to solve more than 50 questions  at least. But in this case, I don't think it can help me get a factor. Sorry if I offend you. But please read the things I posted, and I hope you understand that.
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
And thank you for your time explaining that to me.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
I agree @Callisto u cant get the answer in ths method. I tried with this method at first but didnt get the answr.
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
*have been using that method to solve. Typo. Sorry
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
Ah so you already know synthetic division, I am not offended at all. I just wished to help but unfortunately it's not working. Sorry for causing confusion then.
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
@bronzegoddess Thanks for your time spent here and willingness to help! I appreciate your efforts put here.
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
But there is of 4 factors: option1: 2 real(a+ and a ) and 2imaginary or option2: they are all imaginary isn't it?
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
May be of help @ some other time:)
 one year ago

CallistoBest ResponseYou've already chosen the best response.2
I just need real factors. The expression can be factorised, according to wolf
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
this sorts of questions can be solved without using calculators only by splitting the terms and by adding and removng additional terms. we will knw which terms to add and remove nly by workng many more questions of this type. I dnt thnk there is a specific way to solve this as far as I knw.
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
Sorry about so long attachments
 one year ago

ParthKohliBest ResponseYou've already chosen the best response.0
I'm astonished. A question that looks as easy as simple factorization is confusing the polymaths.
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
attachment is about factorizing:)
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
when you find an answer please do inform me :) thank you!
 one year ago

ajprincessBest ResponseYou've already chosen the best response.1
Bt seems so useful @maheshmeghwal9.
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
thanx:) @ajprincess
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
If u find answer @Callisto Plz inform me:)
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
The general way to solve quartic equation is to Ferrari's method http://www.proofwiki.org/wiki/Ferrari%27s_Method ... for this particular method, assuming it has two quadratic factors, \[ (x^2 + ax + b)(x^2 + cx +d) = x^4  7x^2 + 1 \\ \text{ or, } \\ x^4 + (a+c)x^3 + (b+d+ac)x^2 + (ad+bc)x + bd = x^4  7x^2 + 1 \\ \text{ Comparing coefficients } \\a+c = 0; \\b+d+ac=7 \\ad+bc=0 \\ bd = 1 \] Now solving the values these equations we get the coefficients http://www.wolframalpha.com/input/?i=c%2Ba%3D0%2Cb%2Bd%2Bac%3D7%2Cad%2Bbc%3D0%2C+bd%3D1 Hence we have our quadratic factors http://www.wolframalpha.com/input/?i=factorize+x^4++7x^2+%2B1 The other (numerical) method would be Newton's method http://en.wikipedia.org/wiki/Newton's_method The best rule would be to solve this type of equation simply is view analytically the function f(x) = x^4 7x^2 + 1 eg, f(x) = x^4 +7x^2 +1 does not have any real solutions Besides Descartes's rule sign change and other techniques would be helpful if we were able to guess it's solution (using remainder theorem) then also we can easily factor out using synthetic division
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
@experimentX that was what i was saying all along :) is there a way to find the imaginary solutions then?
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
yes there is, Ferarri's method would be the general method ... check that link it finds out both real and imaginary roots ... (it's a lot messy though ... even for simple problems) Easy method is to reduce it into quadratic factors or, reduce into cubic (xa)(x^3+bx^2+cx+d) = 0 and use Cardano's method < this is messy too (x^2+bx+c)(x^2+dx+e) = 0 < this is the best you can expect from that equation as you know the roots are imaginary if 4ac > b^2 for quadratic factors.
 one year ago

maheshmeghwal9Best ResponseYou've already chosen the best response.0
Hmm.... I heard of Ferrari today "A car also comes of help to maths{LOL}"
 one year ago
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