Question regarding solution to Polynomial

[frog1944]

Hi,

For question 7 c) iii) of 2010 HSC MX2, they state part of the marking criteria to examine the limits as x approaches positive and negative infinity, why do you have to do that?

(http://educationstandards.nsw.edu.au/wps/wcm/connect/7ece6914-bb45-43c6-b035-e4c09ef7eb3e/maths-ext2-hsc-mg-2010.pdf?MOD=AJPERES&CACHEID=ROOTWORKSPACE-7ece6914-bb45-43c6-b035-e4c09ef7eb3e-lGcEkXi, page 10)

Thanks

[RuiAce]

Being realistic, I don't think it's possible for there to be at least two other roots, given that we've checked the stationary points as well. But if we didn't perform this check, the possibility of no other roots is certainly feasible.

Then, there would be no other root, because the curve won't cut the x-axis again.

[frog1944]

That makes sense. But as it's a polynomial (I don't know of any polynomials that as x approaches infinity p(x) doesn't approach infinity of some sort), wouldn't the two stationary points guarantee it has another root?

[RuiAce]

That makes sense. But as it's a polynomial (I don't know of any polynomials that as x approaches infinity p(x) doesn't approach infinity of some sort), wouldn't the two stationary points guarantee it has another root?

Unfortunately, this is basically making the reader assume for themselves the limiting behaviour of a polynomial itself. By stating this, you make it clear.

There's also a subtlety (that I missed earlier) in that the question specified n must be an odd integer. The limiting behaviour would be different if n had been even. It's possible that if n were even, there wouldn't be exactly one other root either.

[frog1944]

Oh, ok. Awesome! Thanks RuiAce

[rickybennett]

How do we solve the following:
P(z)= z^4 - 2z^3 + az - 9
find the value of a if p(1 + i sqr root 2)

[RuiAce]

How do we solve the following:
P(z)= z^4 - 2z^3 + az - 9
find the value of a if p(1 + i sqr root 2)

This question is lacking information. Presumably \(P(1+i\sqrt2)\) is equal to something but you haven’t specified what.