## SheldonEinstein Group Title 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 one year ago one year ago

1. ParthKohli Group Title

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

2. ParthKohli Group Title

Now see what all is true.

3. SheldonEinstein Group Title

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

4. ParthKohli Group Title

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

5. ParthKohli Group Title

Hindi mein bolun?

6. SheldonEinstein Group Title

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

7. SheldonEinstein Group Title

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 Group Title

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

9. ParthKohli Group Title

10. SheldonEinstein Group Title

OK thanks for the shortcut friend...

11. ParthKohli Group Title

|z| is the modulus?

12. SheldonEinstein Group Title

Yeah

13. SheldonEinstein Group Title

It is modulus.

14. ParthKohli Group Title

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

15. SheldonEinstein Group Title

:) Yeah Yeah Yeah!!! Thanks!

16. ParthKohli Group Title

You're welcome ;)