Hello guys, I'd like to ask you how to efficiently factorize complicated polynomial like this one for example: $$\frac{128t^4+128t^3+192t^2+32t+40}{(4t^2+1)^4}$$ I've spend more than a hour trying to decrypt and decompose the polynomial, but to no avail. For simple cubic polynomial I know it might be possible to try ##-1## or ##1## or the factor of the lowest term to get the correct linear factor, which can then be used to do long division to get the other factor if there's any. Eventually I thought it's the end of the line, but then when I checked to WA. the nominator decomposed to this. $$8(4 t^2+1)(4 t^2+4 t+5)$$ How can we know the factor if it's not linear i.e ##(x-a)##? For the note I'm currently reading the Complex Numbers section in Boas book. So far the author has not discussed yet techniques to factor polynomial like this. Thank You
Hello Seydlitz! (btw, it's "numerator" and "denominator" ) I don't understand why you don't simply use long division, it's reasonably quick. Alternatively, if you know that (4t^{2}+1) is a factor, then you can find the other quadratic more or less immediately, by inspection.
As an alternative, the first thing would be to notice that all of the terms in the numerator are even, so bring out a factor of 2. Better yet, notice that the coefficients are all divisible by 8. This means that you can write the numerator as 8(16t^{4} - 16t^{3} + 24t^{2} + 4t + 5). The Rational Root theorem can be used to find roots of the form p/q, where p = ±1 or ±5, and q = ±1, ±2, ±4, ±8 or ±16.
Ouhh my bad. I typed it at night. I know it's possible to use long division because the polynomial is rational, but what if I've only been given the numerator. How can I test quickly that it can be decomposed? If the root is real then I think I can guess it fine, but then in this case the polynomial has complex roots. ##4t^2+4t+5## and the ##4t^2+1##.
I guess the minus sign in the first fraction is a typo? For the factorization, you need a plus there. Well, it has order>2, it has to be composite. To make numbers smaller, I would substitute 2t=s: 8(s^4 + 2s^3 + 6s2 + 2s + 5). You can test imaginary numbers together with real numbers, it will need more time but it is possible. You can even see ±i as a solution here.
(just got up :zzz:) no way that i know of (but don't worry … they won't give you anything that horrible in the exam! )
Yes it should be positive for the second term, sorry for that. I can't no longer edit the first post. Ah so it's possible to check it with imaginary number as well. I'll try to keep that in mind. Thanks for your assurance tiny-tim. It's not everyday I see polynomial like it. I just need a bit of clarification there finally, if a polynomial has order>2 then it has to be composite but it's not necessarily possible to factorize them into simpler factor. Is that correct?
no every real polynomial can be factored into linear factors: (x -a_{1})(x -a_{2})…(x -a_{n}) but the roots may be complex, except: if a is a complex root, then so is a* (its complex conjugate), and so a + a* and aa* are both real, and so (x -a)(x - a*) = x^{2} - (a + a*)x + aa* is real so any real polynomial can be factored into real quadratic factors, plus (if it has odd order) one real linear factor
Ah ok, so every real polynomial can be factored into real quadratic factors or with the addition of one linear factor, the quadratic factors can then be further factorized to complex linear root if it has no real linear root. Is this correct?
If you're only interested in real roots, it's worth noting that we can see instantly that it has no positive real roots, since none of the terms can be negative. If we look at the dominance of the terms for the even coefficients, we should be suspicious that it has no negative real roots either. If you mirror your equation about t=0, by replacing t with -t, then you may be able to convince yourself of this. Your lowest order term is a constant. The next term linear. As the order increases their dominance as we move outside of the range of -1<t<+1 increases. Even terms are always offer a positive contribution and odd terms offer a positive contribution in for t>0 and a negative contribution for t<0. At a guess, for even order polynomials, I would expect there to be a heuristic that suggests there are no real roots where all coefficients are positive; odd coefficients are less than the previous even coefficient and even coefficients are greater than or equal to, the previous odd coefficient. Which matches your equation. It's far from a proof but getting a bit of an intuitive understanding of the problem that you're trying to solve can save you a lot of time. Analytical solutions for the roots of higher order polynomials get very complicated fast. If you need a solution then practically, you would turn to numerical methods.
@craigi: To see that there are no negative real roots, consider the cases -1 < t < 0 and t <= -1: t>-1: 192t^2 + 128t^3 > 0 and 40 + 32t > 0 and 192t^4>0 t<=-1: 192t^4 + 192t^3 > 0 and 192t^2 + 32t > 0 and 40>0