Let a, b, c, d be real numbers, not all zero. Prove that the roots of the polynomial

x^6 + a.x^3 + b.x^2 + c.x + d

cannot all be real.

Here's my method.Let the given polynomial be denoted as

f(x)=x⁶+ax³+bx²+cx+d then

f(-x)=x⁶-ax³+bx²-cx+d

Now,according to descartes rule of change of signs,the max. no. of +ve real roots are no. of sign changes in f(x) and max.no of -ve real roots are no. of sign changes in f(-x).

Now,If a,b,c,d>0 then no sign changes in f(x) and 4 in f(-x).So,A max. of 4 real roots can exist.

If a,b,c>0 and d<0 then 1 in f(x) and 3 in f(-x).So,again a max. of 4 real roots.

Similarly,you can observe that whatever may be the signs of a,b,c,d,You get a max. of 4 real roots to the eqn.So,All the roots of the eqn. can't be real.

f(x) = x^6 + ax^3 + bx^2 + cx +d

ASSUMPTION:all roots are real

now if all 6 roots are real then f(x) becomes 0 @ 6 values of x.. then f'(x) becomes 0 @ 5 pts..f''(x) @ 4 pts and f'''(x) becomes 0 @ 3 REAL  values of x

f'''(x)=120x^3 + 6a which is an inc func and hence it will become 0 only @ 1 val of x..hence our assumption is wrong..

Hsbhatt sir told me that he is having some problem opening this question.When I asked him whether my method was correct,He told it was correct and suggested one more method which is really simple and good.He asked me to post his solution.So,here's the solution suggested by him.

For the given eqn. x⁶+ax³+bx²+cx+d=0,If the roots are x_i(i=1,...6) then

Since a,b,c,d are not all 0,It follows that all the roots are not real.