Bazza's 3/4 Question Thread

[WhoTookMyUsername]

1) In the expansion , the coefficient of the second term is -192. Find the value of n.

Putting this algebraicly i got

I got this to 


My question: is there any way to solve this other than using the CAS or trial and error (is this solvable by hand)?

thanks

don't bother doing this until i can use tex (5 mins)

[moekamo]

well binomial theorem only works for integer values of n as far as your concerned(it works for other values but you don't need to worry about that in year 12 methods), so just plug in values of n and by trial and error you should get n=6, i cant think of any other way to solve that...

[pi]

My question: is there any way to solve this other than using the CAS or trial and error (is this solvable by hand)?

Don't think this is solvable by hand, use CAS. However, it would be easier to sub in integers into the original equation.

[paulsterio]

Do you do a in ascending or descending powers, cause that will affect what the second term is, that's ambiguous :S

[dc302]

Conventionally it should be ascending powers.


edit: although in high school it was always descending powers iirc.

[TrueTears]

Do you do a in ascending or descending powers, cause that will affect what the second term is, that's ambiguous :S

you can do it in whatever order, just find the one with -192 as the coefficient

[Planck's constant]

Even though these questions are meant to be done on CAS, there is a roundabout 'proof' that n=6 (assuming n is integer)

* You can quickly get from where you left off to : 192 = n * 2 ^ (n-1)
* (and noting that 192 = 3 * 2 ^ 6) you can easily get to : 3 = n * 2 ^ (n-7)
* Therefore n is a multiple of 3 and can thus be written as n = 3 * k, and by substitution into the previous line you can quickly arrive at : 2 ^ (7 - 3 * k) = k
* Therefore k is a multiple of 2 and can thus be written as k = 2 * m, which reduces the above equation to : 2 ^ [6 * (1 - m)] = m
* but here m can only = 1 (think about it)
* combining m =1, k = 2 * m, n = 3 * k   => n = 6

 

[dc302]

Even though these questions are meant to be done on CAS, there is a roundabout 'proof' that n=6 (assuming n is integer)

* You can quickly get from where you left off to : 192 = n * 2 ^ (n-1)
* (and noting that 192 = 3 * 2 ^ 6) you can easily get to : 3 = n * 2 ^ (n-7)
* Therefore n is a multiple of 3 and can thus be written as n = 3 * k, and by substitution into the previous line you can quickly arrive at : 2 ^ (7 - 3 * k) = k
* Therefore k is a multiple of 2 and can thus be written as k = 2 * m, which reduces the above equation to : 2 ^ [6 * (1 - m)] = m
* but here m can only = 1 (think about it)
* combining m =1, k = 2 * m, n = 3 * k   => n = 6

 

Just like to add that you should say you've assumed n<7 (or equal) when you claimed that n is a multiple of 3. Since you have n = 3 * 2^(7-n), n is a multiple of 3 if and only if 2^(7-n) is an integer, which is only true if 7-n is non negative.

[WhoTookMyUsername]

What is the best way to factorise polynomials of degree 4 and higher? Through year 11 for degree 3 i was just doing remainder theorem to find one linear factor and then using factorization by inspection to complete it, but this doesn't really work for degree 4 and higher. What is the best way to do it by hand?

[pi]

You'd rarely have to do this tech-free first of all, cubics is usually the highest they go and higher powers become more and more tedious.

The approach is still the same, you find one factor via inspection and then use long division or more inspection to find the rest. If you're lucky, then you may find "grouping" that is present in even powered polynomials, and this may speed things up, but this is rare imo.

Also, you could read up on "synthetic division", which is a bit faster than long division imo.

edit: If you've got a bit of free time and feel like using some hardcore formulas, these might help for cubics and quartics, but personally, I'd stick to more conventional methods

[WhoTookMyUsername]

I'm not sure how to do it that way once i have a linear factor and a cubic factor: how exactly can i factorise by inspection or long division?
This quetsion is in the textbook



so is a factor... -> by inspection


Where exactly do i go from here? Use remainder theorem to factorise the second bracket? or inspection again (separately)
what's the best way here :S ? (lol i know this should be easy but... bio... XD)

thanks

[TrueTears]

factorise the cubic, by inspection x = 2 works

btw if u wanna check out some cool algorithms, check out kronecker's algorithm and http://en.wikipedia.org/wiki/Gauss%27s_lemma_%28polynomial%29 these normally help out when factorising polynomials over integral fields.

[paulsterio]

I wouldn't bother with using the factor theorem again

x^3 + 3x^2 - 4x - 12
x^2(x + 3) - 4(x + 3)
(x^2 - 4)(x+3)
(x-2)(x+2)(x+3)

[pi]

I wouldn't bother with using the factor theorem again

x^3 + 3x^2 - 4x - 12
x^2(x + 3) - 4(x + 3)
(x^2 - 4)(x+3)
(x-2)(x+2)(x+3)

This is the "grouping" I was referring to earlier btw, it doesn't happen all the time though. I would jump to long division (or synthetic division) from there in cases where "grouping" wasn't present (it rarely is btw)

[paulsterio]

This is the "grouping" I was referring to earlier btw, it doesn't happen all the time though. I would jump to long division (or synthetic division) from there in cases where "grouping" wasn't present (it rarely is btw)

in all honesty though, you'd rarely be expected to substitute values and perform long division in an exam 1, even if this was the case, the factors would be something like (x-1)