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

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May be we can take x^2 as "t" & then factor
I mean\[(t^2-7t+1+5)-5\]
If my brain still works, Then it becomes (t-6)(t-1)-5 and then ..?

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Other answers:

I should learn this :/
ya it is not going to be factor like this:/ May be @satellite can help us:)
And he's offline
oops sorry lol
so we made only this much\[(x^2-6)(x^2-1)-5.\] now what should we do of {-5}?
I don't think we should do it in this way...
ya u r right @Callisto But I m thinking more on this.
you need to use synthetic division
@bronzegoddess Can you further explain it?
@ParthKohli It's not like that.... and you made mistakes there too...
I deleted it before you noticed :P
I noticed before you deleted :P
ya i too
This degree 4 is eating my head >.<
I m feeling giddy too
@maheshmeghwal9 can you call Yogesh please?
but he is sitting aside of me
we 3 are all here:)
Haha nice
okay, lets make a guess that (x-2) 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(x-2) isnt a factor because it gives us a remainder 11.
But my wolfy says (x^2 - 3x + 1)(x^2 + 3x + 1)
@bronzegoddess According to the word 'factor', I don't expect there is a remainder.
ya @Callisto is right: D
@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
Hmm but I was telling bronzegoddess that this can be factored. I wanna understand too
Not using synthetic division like that - yes :) Lana~~~~~~~~~~~~
@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!
We must want an exact & good method.
okay, let me do the whole thing again.
@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?
very good @ajprincess
but how did she know about adding -3x^3 & something other. I want to hear from her.
ya but if we didn't know the answer, then what to do It is the exact problem:/
ya will show that in a few minutes time.
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(x-1) and if it also give you a remainder then all your factors are imaginary. Thats how I learnt it.
Do you know when to use that method?
we use trial and error method when we need to factorise expressions with greater power
when looking for factors or zeros of polynomials
@bronzegoddess please take a look at this: We can apply that method when one of the factors is in the form (x-k) Now, in this case, no factors are in the form (x-k). So, it cannot be applied in this case.
Of course you can keep using that method for checking, but you can't get a factor, I think.
@Callisto (x-k) aka (x-1)=0 is x=1 which you'll end up using as the divisor in synthetic division.
Okay, to be simple, if you can get a factor using that method, please tell me.
my step by step explanation from before doesn't make sense?
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.
And thank you for your time explaining that to me.
I agree @Callisto u cant get the answer in ths method. I tried with this method at first but didnt get the answr.
*have been using that method to solve. Typo. Sorry
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.
@bronzegoddess Thanks for your time spent here and willingness to help! I appreciate your efforts put here.
But there is of 4 factors: option1: 2 real(a+ and a -) and 2imaginary or option2: they are all imaginary isn't it?
I just need real factors. The expression can be factorised, according to wolf
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.
Sorry about so long attachments
I'm astonished. A question that looks as easy as simple factorization is confusing the polymaths.
attachment is about factorizing:)
when you find an answer please do inform me :) thank you!
Bt seems so useful @maheshmeghwal9.
thanx:) @ajprincess
If u find answer @Callisto Plz inform me:)
The general way to solve quartic equation is to Ferrari's 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 Hence we have our quadratic factors^4+-+7x^2+%2B1 The other (numerical) method would be Newton's method'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
@experimentX that was what i was saying all along :) is there a way to find the imaginary solutions then?
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 (x-a)(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.
Hmm.... I heard of Ferrari today "A car also comes of help to maths{LOL}"

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