Question

How many polynomials having degree 2009 cannot be factorized:

• Infinitely many
• 2009
• 1
• 0

They should have only real coefficients

Solution

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.

Alternatively

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.