3 unit polynomial question (1 Viewer)

[jsttesting]

Hey guys,

I dunno if this is a simple or hard question, but I cant' seem to get anywhere with trying to solve the following question:


If a, b, c are the roots of the equation x^3 - x^2 + 4x - 1 = 0, find the value of (a+1)(b+1)(c+1).

Please let me know if you get the answer.....

Thanks...!

 

[KFunk]

(a +1)(b +1)(c +1)
=(ab +a +b +1)(c +1)
= abc +(ab +bc +ca) + (a + b + c) +1
then using the values gained from summing the roots of P(x)= x³ -x² +4x - 1
= 1 +(4) + (1) +1
=7

 

[jsttesting]

Ahhh.......I get it now.....thanks heaps...!!

 

[KFunk]

No problem .

 

[Slidey]

Alternatively:

roots a+1, b+1, c+1, then y=x+1, x=y-1
sub into:
x^3 - x^2 + 4x - 1 = 0
to get:
(y-1)^3 - (y-1)^2 + 4(y-1) = 1
y^3-3y^2+3y-1-y^2+2y-1+4y-4-1=0
y^3-4y^2+9y-7=0

product of roots = -d/a = 7 = (a+1)(b+1)(c+1).

 

[JamiL]

slide rule is better, less room 4 error... or rather the 1 i would of use n came 2 my head 1st

 

[KFunk]

Came to my head first as well but it's the 3U forum and I just tend to think of that as a 4U technique. I don't know about less room for error though...

 

[Trev]

Came to my head first as well but it's the 3U forum and I just tend to think of that as a 4U technique. I don't know about less room for error though...

Agreed, the first method used I find to have less chance of error. Expanding 'tings' such as (a+1)(b+1)(c+1) generally come out in groups/patterns, so it is easy to tell whether you have screwed up the working.

 

[JamiL]

yer when i first saw it and it came in2 my head i wasn sure wether it was 4 or 3u?? but i find it easyer and less rom 4 error... althou it depends on da person...
the expantion of (a+1)(b+1)(c+1) looks way 2 messy 4 my likeing, n its easy 2 expand (x-1)^3 but expanding (a+1)(b+1)(c+1) u need 2 do it in 2-3 steps...
i dunno... y do u think the 1st method was better?

 

[KFunk]

Just because, in terms of an explanation, it makes more sense. You can see directly why it works and it's an easy expansion. The technique of creating a polynomial with new roots is far more subtle.

 

[JamiL]

far more subtle 2 a person who hasnt learnt it... lol.. althou even u said it came 2 ur head 1st and me and slide rule did it similarly, so it can be that subtle. in sayin that both would get u full marks and i would use the 1 that u like better... so i fink this argument is futile... lol

 

[Slidey]

I think there's MORE room for error with my method. Suppose it asked for a^2+b^2+c^2 instead? Square roots!

 

[JamiL]

again 2 methods there slide rule..
u could go the conventional (a+b+c)^2 - 2(ab + bc +ca) which i would use in this case
or u could use
a=X^(.5) n create new polynomial... again up 2 the person, and in this case it makes it a little more difficult but stil not impossible... i tried it n that method would be a stupid 2 do cos ur dealin with x^3 n ull end up with X^(1.5) n u have 2 re arrange it n stuff, not worth it....