HSC Tips - Polynomials (1 Viewer)

[abdooooo!!!]

ah yes hsc science... the pinnacle of all logical thoughts.

 

[McLake]

Originally posted by abdooooo!!!
syllabus??? who the hell reads that?

are they gonna give you a question like "explain qualitatively the difference between roots and zeros?"

Yes, they possibly could.

I have seen questions that ask you to state de Moivre's, or why a complex poly has conjugate roots (assuming real coefficents), so why not ask that?

 

[freaking_out]

Originally posted by McLake
Yes, they possibly could.

I have seen questions that ask you to state de Moivre's, or why a complex poly has conjugate roots (assuming real coefficents), so why not ask that?

yeah, i've seen a few "explain" answer as well in 3u trials.

 

[KeypadSDM]

Originally posted by McLake
Yes, they possibly could.

I have seen questions that ask you to state de Moivre's, or why a complex poly has conjugate roots (assuming real coefficents), so why not ask that?

Thank God I'm out of there.

If the maths course derides into that ... Ugh, I can only hope for the future mathematicians this state produces.

 

[freaking_out]

Originally posted by KeypadSDM
Thank God I'm out of there.

If the maths course derides into that ... Ugh, I can only hope for the future mathematicians this state produces.

but those questions are not as bad as the questions u get in hsc science.

 

[KeypadSDM]

Originally posted by abdooooo!!!
are they gonna give you a question like "explain qualitatively the difference between roots and zeros?"

Originally posted by McLake
I have seen questions that ask you to state de Moivre's, or why a complex poly has conjugate roots (assuming real coefficents), so why not ask that?

It'd be so much easier if the question asked QUANTITATIVELY. Then you could just say 0. Easy.

 

[freaking_out]

Originally posted by KeypadSDM
It'd be so much easier if the question asked QUANTITATIVELY. Then you could just say 0. Easy.

what do u mean?

 

[budj]

Umm.. I think in the syllabus it states somewhere that we need to prove the fundamental theorem of algebra. I think it is the embodiment of this statement
" The fundamental theorem of algebra asserts that every polynomial of degree n over the complex field has at leastone root. Using this result, the factor theorem should now be used to prove (by induction on the degree) that a polynomial of degree n>0 with real (or complex) coeffeicients has exactly n complex roots (each counted according to its multiplicity) and is expressible as a product of exactly n complex linear factors.

So theoretically they can ask a question which states prove the fundamental theorem of algebra, in a 4 unit exam yeah?

 

[zergcave]

man..polynomials seems like a topic as tricky as complex numbers...

can somone plz provide solutions for fitzpatrick 4u - ex 36c. questions 2 - 10

 

[Slidey]

I will tomorrow or something when I'm meant to be doing some mind-numbingly boring and useless task for homework.

 

[KFunk]

man..polynomials seems like a topic as tricky as complex numbers...

can somone plz provide solutions for fitzpatrick 4u - ex 36c. questions 2 - 10

Here's a couple to start with, I'll type up c) and d) if I feel game but long expansions with lots of powers are hell.

2. ax⁴ + bx³ + cx² + dx +e =0

where α + β +γ + δ = -b/a
αβγδ=e/a

a) P(x/m) = am^-4x⁴ + bm^-3x³ + cm^-2x² + dmx +e

where α + β +γ + δ = -(b/m³)/(a/m⁴) = -(b/a).m (hence P(x/m)=0 has roots m times those of P(x)=0)

b) P(1/x) = ex⁴ + dx³ + cx² + bx +a
αβγδ=a/e = (e/a)^-1 (hence the roots of P(1/x)=0 are the reciprocals of those of P(x)=0)

 

[KFunk]

3.
(cosθ + isinθ )⁴ = cos4θ + isin4θ
or
(cos&theta; + isin&theta; )⁴ = cos⁴&theta; +4icos<su>3</sup>&theta; sin&theta; -6cos<eup>2</sup>&theta; sin²&theta; -4icos&theta; sin³&theta; +sin⁴&theta;

Then, equating real parts.

cos4&theta; = cos⁴&theta; -6cos<eup>2</sup>&theta; sin²&theta; +sin⁴&theta;
= cos⁴&theta; -6cos<eup>2</sup>&theta; (1- cos²&theta; ) + (1 - cos²&theta; )²
= 8cos⁴&theta; - 8cos²&theta; +1

So if you substitute x=cos&theta; into:
8x⁴ -8x² +1 = 0 then it shares solutions with cos4&theta; = 0

==> 4&theta; = &pi;/2, 3&pi;/2, 5&pi;/2, 7&pi;/2 ===> &theta; = &pi;/8, 3&pi;/8, 5&pi;/8, 7&pi;/8

roots of the equation are cos&pi;/8, cos3&pi;/8, cos5&pi;/8, cos7&pi;/8

a) summing roots one at a time you find that
cos&pi;/8 + cos3&pi;/8 + cos5&pi;/8 + cos7&pi;/8 = 0

b)summing roots 4 at a time you find that:
cos&pi;/8cos3&pi;/8cos5&pi;/8cos7&pi;/8 = 1/8

 

[undalay]

k i got a weeny lil question...but wuts the difference between the roots of a polynomial and the zeros of a polynomial?
thanking u muchly

Find the roots of x^2 + x = 4
is the same as
Find the zeroes of x^2 + x - 4


Find the zeros of x^2 + x
is the same as
Find the roots of x^2 + x = 0

I'm pretty sure this is the difference, although could be more to it.

 

[wogblogger]

hmmmmmm
not sure
but by definition a root of a polynomial is the point(s) it crosses the x-axis

soo by applying bs to this
it can be assumed a zero of a polynomial is when the "function" has a zero value

farout can 4unit stop trying to be adv.english

mmmmm.......

(year 3 maths) + (modern history 2unit) = Physics 2unit

(year 3 maths) + (modern history 2unit) + (adv.english essay skills)
=
1st in state Physics 2unit

......lol sorry about my outburst

 

[kevinant]

- Know how to do long divsion with polys, complex polys, unknonw coeffecient polys.

Synthetic Division would be MUCH FASTER, EASIER and TAKE LESS SPACE
I reckon that's one of the key to finish the whole paper without missing any questions!

EXPLAIN why 3 + i is a root

ANSWER: "If a polynomial with REAL coefficients has a complex root, the its conjugate is also a root." (This was in last years HSC)

I actually wrote lots on this question... not just stating that fact but I also went on with it has to be conjugate otherwise it will result in a complex coefficient and stuffs like that... I don't know if that was extra thing and wasting time....