polynomials question (1 Viewer)

[u-borat]

but the relationship between the coefficients stays the same.

 

[Volt]

its equal to a^2(x - 1)(x - 4) not (x - 1)(x - 4)

eg. 5(x - 1)(x - 4) is divisible by both (x - 1) and (x - 4)

I interpreted exactly divisible (even though I made no mention of 'exactly' in my post - I should have - but it is a small omission) in the same sense we could say that 6 is exactly divisible by 2 and 3: only 1 is left when when dividing by both. But in any case it is probably the wrong interpretation given the wording of the question.

But it doesn't invalidate the far simpler course of inquiry (or apparently simpler, I can't be bothered reading the other solutions) even if we assume the greater generality. The solution can be easily modified to account for this, and it so happens that the answer is the same (the greater generality doesn't yield any extra solutions).

 

[Volt]

this is an incorrect definition of exactly divisible (heres some examples i could find on the net: example questions)

if (x - b) exactly divides P(x), P(b) = 0, this is the true mathematical definition.

(x - b) can divide P(x), however P(b) could be equal to the remainder.



P(x) = (ax + b)^2 - x

= a^2 x^2 + 2axb + b^2 - x

= (a^2)x^2 + (2ab - 1)x + b^2

note the coefficient on the x^2, if you dont consider this on your RHS (i.e. RHS = a^2(x - 4) (x - 1) ) you immediately assume a = 1 or a = -1, just so happens this is the case with this question!

Hehe, true. I acknowledged this in my post.

But it is clearly the better solution rather than tediously equating all coefficients and considering simultaneous equations, etc., as some of the posts here. So more properly:

We want:

(ax + b)^2 - x ≡ { (x - 2)^2 - x }c for some integer constant c (arrived to as before).

It is easy to see that c must equal 1, and you simply conclude as before. In an exam sort of situation it would be vastly quicker to write this out without error, it would be just a few lines and much more direct...

Wow. So much discussion for such a simple question!

 

[lyounamu]

Can I quickly ask question here:

Find the number of ways arranging 10 ladies and 10 gentleman around a table:

19! (I got this)

Given that sue, john and ray are amongst these 20 people find the probabilty that sue will sit between two.

 

[3unitz]

Can I quickly ask question here:

Find the number of ways arranging 10 ladies and 10 gentleman around a table:

19! (I got this)

Given that sue, john and ray are amongst these 20 people find the probabilty that sue will sit between two.

treat the 3 as 1 person so the total arrangements will be 17! and also include the 2 possibilities:

1) john, sue, ray
2) ray, sue, john

so the total number of ways = 17! x 2

probability = 17! x 2 / 19! = 1/171

 

[lyounamu]

treat the 3 as 1 person so the total arrangements will be 17! and also include the 2 possibilities:

1) john, sue, ray
2) ray, sue, john

so the total number of ways = 17! x 2

probability = 17! x 2 / 19! = 1/171

Awesome.

 

[shaon0]

Yea equating out the coefficients is a little quicker....thats how i didn't do it.

No, my bad i didn't equate the co-efficients.
I just used simultaneous equations and i did it again using lyounamu's way.
Easy question even though i am not qualified in 4unit maths.