How many polynomials having degree 2009 cannot be factorized:
- Infinitely many
They should have only real coefficients
This question has can be solved in atleast 2 ways.
Every polynomial is continuous function.
So if we can find 2 !!x!! such that !!p(x)!! is positive at one and negative at another. Then we can say !!p(x)!! will have atleast one real root.
The polynomial !!p(x)!! will of the form !! k_0x^2009+k_1x^2008+k_2x^2007+...+k_2009!!
If !!k_0>0!! as !!x->oo!!, !!p(x)->oo!! and as !!x->-oo!!, !!p(x)->-oo!!
If !!k_0<0!! as !!x->oo!!, !!p(x)->-oo!! and as !!x->oo!!, !!p(x)->oo!!.
Since !!p(x)!! has opposite signs at !!-oo,oo!! so it will have atleast one root.
As many have suggested, if !!z!! is a root of a polynomial !!p(x)!! then !!barz!! is also a root of the equation.
As the given polynomial is of degree !!2009!! and suppose all the roots are complex. Since there are odd number of roots. So for atleast one root !!z=barz!!. Which implies !!z!! is real, this is a contradiction to our assumption.
So all roots of the polynomial can not have be complex.
Let us prove that if z is a root of polynomial, !!barz!! is also a root of the polynomial.
Now !!z=ke^(itheta)!!, then !!z^a= k^a e^(aitheta)!!, !!barz=ke^(-itheta)!!, then !!(barz)^a= k^a e^(-aitheta)!!
!!Re(barz)^a= Re z^a!! and !!Img(barz)^a=-Img z^a!!
So we separate real the imaginary and real part. We will get !!p(z)=0+i0!!, so the !!p(barz)=0-i0!!. Hence !!barz!! is also a root of the equation.