## SheldonEinstein 3 years ago 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

1. ParthKohli

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

2. ParthKohli

Now see what all is true.

3. SheldonEinstein

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

4. ParthKohli

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

5. ParthKohli

Hindi mein bolun?

6. SheldonEinstein

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

7. SheldonEinstein

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

8. SheldonEinstein

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

9. ParthKohli

10. SheldonEinstein

OK thanks for the shortcut friend...

11. ParthKohli

|z| is the modulus?

12. SheldonEinstein

Yeah

13. SheldonEinstein

It is modulus.

14. ParthKohli

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

15. SheldonEinstein

:) Yeah Yeah Yeah!!! Thanks!

16. ParthKohli

You're welcome ;)