how factor a polynomial into 2 quadratics?

hi I have a 4th order polynomial like so

mypolynomial = x^4 +a*x^3 + b*x^2 + c*x + d == 0

I now want to factorise this into 2 quadratic equations, as either quadratic equation contains the natural frequency and damping of a certain flight mode of an aircraft. Is there a way to do this? I have tried the 'factor' function, and also tried

collect(mypolynomial, x^2)

however this just returns the original fourth order polynomial. The equation i wish to obtain looks something like this

(x^2 + A*x + B)(x^2 + C*x + D) = 0

where the upper case coefficients are not the same as the lower case coefficients in 'mypolynomial'.

Any help is appreciated, thanks!

Answer 1

[John D'Errico]

You cannot uniquely factor a 4th degree polynomial into such a pair of quadratics. You may think that you can, but it is provably impossible to do so, and a simple counter-example is sufficient to show why.

syms x
quartic = expand((x-1)*(x-2)*(x-3)*(x-4))

quartic = 

But that polynomial can be trivially written as the product of two quadratic polynomials.

Q1 = expand((x-1)*(x-2))

Q1 = 

Q2 = expand((x-3)*(x-4))

Q2 = 

expand(Q1*Q2)

ans = 

As you can see, Q1*Q2 must yield the same fourth degree polynomial. But is there any reason I could not have done this?

Q1 = expand((x-1)*(x-3))

Q1 = 

Q2 = expand((x-2)*(x-4))

Q2 = 

expand(Q1*Q2)

ans = 

So we have completely different quadratic factors. There is indeed no unique way to write such a 4th degree polynomial. This is no different from saying that an integer like 210 = 2*3*5*7, can be written in any of the forms 6*35 = 10*21 = 15*14. There is no unique factorization possible. The same idea applies to polynomials.

All that you can do is to find all 4 roots, then you could pair them up in any order you wish, Whatever makes you happy. This would suffice:

xroots = solve(mypolynomial,'maxdegree',4);

Or, if the coefficients of your polynomial are all numerical values, then you can use vpasolve.

xroots = vpasolve(mypolynomial);

In some cases, your roots MAY pair naturally up into pairs of complex conjugate roots. But that still does not give you a unique factorization.

[Robert Wake]

hi john, thanks for your answer. Yes, you're absolutely correct for what it is that i wrote. However I missed out a huge bit of info that I think would change things. My polynomial is supposed to be equal to zero. I will edit my post to show this. Is this still the case? Thank you.

[Walter Roberson]

That does not change anything: John's example polynomial equals zero.

In the restricted case that the original coefficients are real valued and the two target polynomials have real coefficients but complex roots, then you might get a unique factorization (if you mismatch the conjugate pairs then they multiply to give complex coefficients rather than real coefficients)

[Robert Wake]

Hi walter thanks I mustn't have been explaining it correctly as I've found a way.

syms x
A = [-0.00687 0.01395 0 -9.81; -0.0905 -0.31498 236 0; 3.89e-4 -0.00336 -0.4282 0; 0 0 1 0];
I = eye(4)  %identity matrix
eqn = det(I*x - A)

where eqn returns the following 4th order polynomial;

eqn = x^4 + 0.75*x^3 + 0.934*x^2 + 0.00945*x + 0.00419

i then use the factor function

quad = factor(eqn, x, 'FactorMode', 'real')

this then returns the equation

quad = [x^2 + 0.00658*x + 0.00453, x^2 + 0.743*x + 0.925]

where the coefficients 0.00453 and 0.925 correspond to the natural frequency of two separate flight modes of a disturbed aircraft. Thanks!

[Walter Roberson]

factor() with 'real' sounds good.