4U Maths Question Thread

[itssona]

yo Rui can u tell the ext 2 methods? XD
find the values of a and b such that x^4 -x^3 +x^2 +ax +b is divisible by (x^2+4)

[RuiAce]

yo Rui can u tell the ext 2 methods? XD
find the values of a and b such that x^4 -x^3 +x^2 +ax +b is divisible by (x^2+4)

where \( P(x) \) is the polynomial you were given, i.e. \( P(x) = x^4 - x^3 + x^2 + ax + b \)

[julies]

hey not sure if this question's been asked already
but I don't understand how to do this q and the solutions don't make sense
thanks in advance!

[RuiAce]

(Image removed from quote.)

hey not sure if this question's been asked already
but I don't understand how to do this q and the solutions don't make sense
thanks in advance!

[Shlomo314]

Hey,
What are the best strategies for the next 3 days to maximise marks before the Ext 2 paper apart from past HSC papers?
Thankyou

[RuiAce]

Hey,
What are the best strategies for the next 3 days to maximise marks before the Ext 2 paper apart from past HSC papers?
Thankyou

I don't know of any better strategy (until either the night before the exam, where you should be resting, and the morning of the exam, where you should just reconsolidate all the methods/formulas/... you know).

If you don't feel like studying for maths anymore (ran out of papers, genuinely bored etc.) then it might be time to study for something else.

Basically, you stop worrying about past papers once you're at the last stretch. Until then, there is still no optimal strategy aside from them.

[itssona]

hiii
so if we know z^3=1
how do we evaluate (1+2w+3w^2)(1+2w^2+3w)

whats the way we do this?

[RuiAce]

hiii
so if we know z^3=1
how do we evaluate (1+2w+3w^2)(1+2w^2+3w)

whats the way we do this?

I’m about to drive home but for reference, did you mean w^3=1?

[beau77bro]

ii and iii, it says "recorded", why don't we care about the arrangements in the answer, specifically ii?

also why can't we do iii, by going 1-probability of a draw and divide by 2?

[itssona]

I’m about to drive home but for reference, did you mean w^3=1?

ohh yeah I mean w^3 sorry

[RuiAce]

ohh yeah I mean w^3 sorry

__________________________________

\begin{align*}\therefore (1+2w+3w^2) (1+2w^2+3w) &= (w^2-1)(w-1)\\ &= w^3 - w^2 - w + 1\\ &+ w^3 - (w^2 - w - 1) + 2\\ &= 1 - 0 + 2\\ &= 3\end{align*}

[RuiAce]

ii and iii, it says "recorded", why don't we care about the arrangements in the answer, specifically ii?

also why can't we do iii, by going 1-probability of a draw and divide by 2?

___________________________

If you mean [1 - Pr(whole match is drawn) ] / 2 then yes you can. But if you do that, just make sure you've covered every single case for a draw.

Terry Lee's answers just chose not to do it.

[beau77bro]

___________________________

If you mean [1 - Pr(whole match is drawn) ] / 2 then yes you can. But if you do that, just make sure you've covered every single case for a draw.

Terry Lee's answers just chose not to do it.

then don't we need to consider the arrangements in the "total number of arrangements"? or something - otherwise isnt it just the probability of a win, law, draw, draw in any order?

[RuiAce]

then don't we need to consider the arrangements in the "total number of arrangements"? or something - otherwise isnt it just the probability of a win, law, draw, draw in any order?

Why does the total number of arrangements matter?

What we've calculated is the probability of W-D-L-D in that order. If we multiply by the number of ways we can arrange WDLD like we did in part iii), then we would have the probability of WDLD in any order and not just that specific one.

[Shlomo314]

Hey, how would I do this question?

Cheers