Rank 4 Unit Topics from easiest to hardest (1 Viewer)

[gcchick]

The questions I remember from poly were almost always the easiest.

 

[the-derivative]

Lol, are you guys aware that the title says 'easiest to hardest'?

Most people don't give a shit.

 

[youngminii]

Whaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaat! The polys in mine took so much time ==;;
There was one question like:
x^4 + 2x^3 + 3x^2 + 2x + 1 = 0
w is a complex cube root of unity.
i) Show w is a root.
ii) Show wconjugate is also a root.
iii) Find the other quadratic factor.
iv) Hence find all roots to the equation.

This was worth 10/65 in a 55 minute exam. Jesus christ how the hell do we have enough time?
Oh yeah I made a mistake in this and I was looked at the clock and I had VERY little time left and I wasn't up to Integration section yet fuckfuck poor time management = fail.

 

[gcchick]

WTF? We never had cube roots of unity in polynomials, as far as I can remember. They were only in complex numbers.

Eh so glad I've graduated.

 

[GUSSSSSSSSSSSSS]

Whaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaat! The polys in mine took so much time ==;;
There was one question like:
x^4 + 2x^3 + 3x^2 + 2x + 1 = 0
w is a complex cube root of unity.
i) Show w is a root.
ii) Show wconjugate is also a root.
iii) Find the other quadratic factor.
iv) Hence find all roots to the equation.

This was worth 10/65 in a 55 minute exam. Jesus christ how the hell do we have enough time?
Oh yeah I made a mistake in this and I was looked at the clock and I had VERY little time left and I wasn't up to Integration section yet fuckfuck poor time management = fail.

poly's q in my half yearlies by far and away the HARDEST on the paper lol

 

[Bacchus]

Whaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaat! The polys in mine took so much time ==;;
There was one question like:
x^4 + 2x^3 + 3x^2 + 2x + 1 = 0
w is a complex cube root of unity.
i) Show w is a root.
ii) Show wconjugate is also a root.
iii) Find the other quadratic factor.
iv) Hence find all roots to the equation.

This was worth 10/65 in a 55 minute exam. Jesus christ how the hell do we have enough time?
Oh yeah I made a mistake in this and I was looked at the clock and I had VERY little time left and I wasn't up to Integration section yet fuckfuck poor time management = fail.

i) w^3 = 1
(w-1)(w^2+w+1) = 0
so w^2 + w + 1 = 0
(w^2+w+1)^2 = 0
w^4 + w^3 + w^2 + w^3 + w^2 + w + w^2 + w + 1 = w^4 + 2w^3 + 3w^2 + 2w + 1 = 0

so since w^4 + 2w^3 + 3w^2 + 2w + 1 = 0,
w is the root.

ii)

let conjugate be cw

(cw)^4 + 2(cw)^3 + 3(cw)^2 + 2(cw) + 1 = c(w^4+2w^3+3w^2+2w+1) = c(0) = 0

iii) just solve w^2 + w + 1 = 0,
iv) keep going. lol

 

[youngminii]

Lol wtf @ your ii)
I used the fact that wconjugate = w^2 when it's a complex cube root of unity. Although I didn't prove that fact, since ii) was only 2 marks.

Also huh @ iii), how does that help?

Anyway, my point was that it took way too much time. (And I made a mistake..)

Btw, who are you?

 

[youngminii]

WTF? We never had cube roots of unity in polynomials, as far as I can remember. They were only in complex numbers.

Eh so glad I've graduated.

Ye lol, I full studied for roots of unity in the first exam (which had complex). Nothing came up with it. Lame <.<

 

[Bacchus]

Lol wtf @ your ii)
I used the fact that wconjugate = w^2 when it's a complex cube root of unity. Although I didn't prove that fact, since ii) was only 2 marks.

Also huh @ iii), how does that help?

Anyway, my point was that it took way too much time. (And I made a mistake..)

Btw, who are you?

ii) is a typical prove question. Refer to cambridge first chapter question on complex.

I am namu btw

for iii) it's just that you have equation given = (w^2+w+1)^2

so you will have 2 equal roots with 2 equal conjugate roots.

 

[GUSSSSSSSSSSSSS]

ii) is a typical prove question. Refer to cambridge first chapter question on complex.

I am namu btw

for iii) it's just that you have equation given = (w^2+w+1)^2

so you will have 2 equal roots with 2 equal conjugate roots.

yeh ii) work out fine

but i still dun see why ur tip for (iii) helps lol..

 

[GUSSSSSSSSSSSSS]

ii) is a typical prove question. Refer to cambridge first chapter question on complex.

I am namu btw

for iii) it's just that you have equation given = (w^2+w+1)^2

so you will have 2 equal roots with 2 equal conjugate roots.

no wait dun worry

it all works out lol

..

but thats sort of longer than other ways to do it..

 

[Bacchus]

yeh ii) work out fine

but i still dun see why ur tip for (iii) helps lol..

you are solving the equation.

solving w^2+w+1 = 0

gives you two roots (normal and conjuage of that)

but since the equation that is given = (w^2+w+1)^2,
you will have 2 roots that are equal and 2 equal conjugate ones.

the roots are (-1 + sqrt(3)i)/2, (-1 + sqrt(3)i)/2, (-1 - sqrt(3)i)/2 and (-1 - sqrt(3)i)/2

 

[Bacchus]

no wait dun worry

it all works out lol

..

but thats sort of longer than other ways to do it..

nah, it isn't

solving w^2+w+1 = 0 takes 30 seconds.

 

[GUSSSSSSSSSSSSS]

nah, it isn't

solving w^2+w+1 = 0 takes 30 seconds.

hahaha i just realised that u werent already given a root

lol i thought u were

then multiplying by conjugate and then doing a synthesis division would take like 10seconds

but lol just looked back and saw u OBVIOUSLY cant do that lol

and yeh so ur way work out best lol

 

[Drongoski]

Re: gcchick's posting


Many hate mechanics simply because they do not have the basic background in statics/dynamics .. kind of stuff that have been taught in the British system from way back. If you do physics for e.g. then much of mechanics would be easier. I suspect many of the younger maths teachers themselves do not have a background in mechanics to handle this topic properly.

 

[youngminii]

Oh sheet it's namu
HOLY CRAP @ YOUR METHOD FOR III), I so did not see that. I made the assumption that w = cis2pi/3 or something, which is fail, I know. Sigh..

Btw,

let conjugate be cw

(cw)^4 + 2(cw)^3 + 3(cw)^2 + 2(cw) + 1 = c(w^4+2w^3+3w^2+2w+1) = c(0) = 0

Can you explain this to me? I'm confused.

 

[Bacchus]

Oh sheet it's namu
HOLY CRAP @ YOUR METHOD FOR III), I so did not see that. I made the assumption that w = cis2pi/3 or something, which is fail, I know. Sigh..

Btw,
Can you explain this to me? I'm confused.

There is a conjugate rule like this:

2cw = c(2w)

cw x ck = c(wk)

cw + ck = c(w+k)

so if you have something like:

(cw)^2 + (cw) + 1

that becomes c(w^2+2+1)

 

[GUSSSSSSSSSSSSS]

There is a conjugate rule like this:

2cw = c(2w)

cw x ck = c(wk)

cw + ck = c(w+k)

so if you have something like:

(cw)^2 + (cw) + 1

that becomes c(w^2+2+1)

lol im sure u (youngminii) will have fun remembering that....lol

 

[youngminii]

Oh is c representing the bar on top? I see, lol that was simple and I didn't even see it, I feel like killing myself.

I made assumption that cw = w^2 (as it's a complex cube root of unity)

 

[waxwing]

I don't know if it helps anybody, but in that question you wrote for part (ii) "Show that w conjugate is also a root".
In part (i) you proved (hopefully) that w is a root, thus there is really nothing to do for part (ii). Since the coefficients of the polynomial are all real, it is known that the non-real roots occur in complex conjugate pairs.

Of course the reason for that principle is just as you've been discussing:


Each of those steps can be justified by observing, geometrically, that the process of reflection is commutative with another geometrical transformation ; dilation, rotation etc.
It's pretty weird but quite cool when you understand it