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

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maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0May be we can take x^2 as "t" & then factor

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0I mean\[(t^27t+1+5)5\]

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

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0I should learn this :/

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0ya it is not going to be factor like this:/ May be @satellite can help us:)

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0so we made only this much\[(x^26)(x^21)5.\] now what should we do of {5}?

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.2I don't think we should do it in this way...

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0ya u r right @Callisto But I m thinking more on this.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you need to use synthetic division

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.2@bronzegoddess Can you further explain it?

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0x^26 @ParthKohli

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.2@ParthKohli It's not like that.... and you made mistakes there too...

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0I deleted it before you noticed :P

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.2I noticed before you deleted :P

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0This degree 4 is eating my head >.<

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0I m feeling giddy too

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0@maheshmeghwal9 can you call Yogesh please?

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0but he is sitting aside of me

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0we 3 are all here:)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay, 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.

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0But my wolfy says (x^2  3x + 1)(x^2 + 3x + 1)

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.2@bronzegoddess According to the word 'factor', I don't expect there is a remainder.

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0ya @Callisto is right: D

Callisto
 4 years ago
Best 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

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm but I was telling bronzegoddess that this can be factored. I wanna understand too

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0LANA THE PROFESSOR :D

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.2Not using synthetic division like that  yes :) Lana~~~~~~~~~~~~

anonymous
 4 years ago
Best 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!

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0We must want an exact & good method.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay, let me do the whole thing again.

Callisto
 4 years ago
Best 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?

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0very good @ajprincess

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0but how did she know about adding 3x^3 & something other. I want to hear from her.

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0ya but if we didn't know the answer, then what to do It is the exact problem:/

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.1ya will show that in a few minutes time.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0first: 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.

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.2Do you know when to use that method?

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.1we use trial and error method when we need to factorise expressions with greater power

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0when looking for factors or zeros of polynomials

Callisto
 4 years ago
Best 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.

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

anonymous
 4 years ago
Best 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.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my step by step explanation from before doesn't make sense?

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.2Sorry, 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.

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.2And thank you for your time explaining that to me.

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

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.2*have been using that method to solve. Typo. Sorry

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah 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.

Callisto
 4 years ago
Best 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.

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

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0May be of help @ some other time:)

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.2I just need real factors. The expression can be factorised, according to wolf

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.1this 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.

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry about so long attachments

ParthKohli
 4 years ago
Best ResponseYou've already chosen the best response.0I'm astonished. A question that looks as easy as simple factorization is confusing the polymaths.

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0attachment is about factorizing:)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0when you find an answer please do inform me :) thank you!

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.1Bt seems so useful @maheshmeghwal9.

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0thanx:) @ajprincess

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0If u find answer @Callisto Plz inform me:)

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2The 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

anonymous
 4 years ago
Best 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?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2yes 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.

maheshmeghwal9
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm.... I heard of Ferrari today "A car also comes of help to maths{LOL}"
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