The polynomial $P(X) = a_0 + a_1X + a_2X^2 + ....+ a_nX^n$ where $a_0,a_1,...,a_n$ are all real numbers and $a_0a_1a_2....a_n0$. Then which of the following MUST be true: P(X) always has a real root P(X) has a complex root z and |z| ≥1 P(X) has a complex root and |z|

Question

The polynomial $P(X) = a_0 + a_1X + a_2X^2 + ....+ a_nX^n$  where $a_0,a_1,...,a_n$ are all real numbers and $a_0>a_1>a_2>....>a_n>0$. Then which of the following MUST be true:

P(X) always has a real root

P(X) has a complex root z and |z| ≥1

P(X) has a complex root and |z|<1

P(X) does not have any complex root

parthkohli · Answer

Always assume a polynomial of this form. Use this for the experiments:$$\rm p(x) = x^2  + 2x +  3$$

parthkohli · Answer

Now see what all is true.

anonymous · Answer

@ParthKohli Sorry, didn't get you, can you explain me more?

parthkohli · Answer

In these questions, I always take an example of a polynomial and see what all is true amongst the choices.

parthkohli · Answer

Hindi mein bolun?

anonymous · Answer

OK... i am getting it... let me check the choices now.

anonymous · Answer

At least it will be complex root if I take your equation of the polynomial. 

P(X) = $x^2+2x+3$ 

x = $\LARGE\frac{-2 \pm \sqrt{4-12}}{2} = \frac{-2 \pm 2\sqrt{-2}}{2} = -1 \pm \sqrt{2} i $

anonymous · Answer

So it will be ,,,, 2nd option :   P(X) has a complex root z and |z| ≥1

Right?

parthkohli · Answer

Keep taking such equations and you'd get your answer... I think.

anonymous · Answer

OK thanks for the shortcut friend...

parthkohli · Answer

|z| is the modulus?

anonymous · Answer

Yeah

anonymous · Answer

It is modulus.

parthkohli · Answer

$$\sqrt{(-1)^2 + (\sqrt2)^2} = \sqrt{1 + 2} = \sqrt 3$$Yes, you're right

anonymous · Answer

:) Yeah Yeah Yeah!!! Thanks!

parthkohli · Answer

You're welcome ;)