Polynomials Question (1 Viewer)

[echelon4]

if a,b,c are the roots of x^3-2^2-3x+1, evaluate

1. a^3+b^3+c^3
2. a^4+b^4+c^4



thanks in advance

 

[SonyHK]

if a,b,c are the roots of x^3-2^2-3x+1, evaluate

1. a^3+b^3+c^3
2. a^4+b^4+c^4



thanks in advance

p(x)=x^3-x^2-3x+1

since a,b,c are roots of p(x).. then p(a)=0, p(b)=0, p(c)=0
a^3-a^2-3a+1=0-------------->(1)
b^3-b^2-3b+1=0-------------->(2)
c^3-c^2-3c+1=0--------------->(3)

(1)+(2)+(3)=a^3+b^3+c^3-(a^2+b^2+c^2)-3(a+b+c)+3=0
u can get a^2+b^2+c^2=... from sum of roots a+b+c=...

times a to equation (1), b to equation (2) and c to equation (3) then add them again to get da right stuffs for the next part

 

[Raginsheep]

2. For this question, if a,b,c are the roots of the equation, then sub in a,b,c
.:. a^3-2^2-3a+1....(1)
b^3-2^2-3b+1.....(2)
c^3-2^2-3c+1.....(3)

(1)x a: a^4-a2^2-3a^2+a=0....(4)
(2)x b: b^4-a2^2-3b^2+b=0....(5)
(3)x c: c^4-a2^2-3c^2+c=0......(6)

(4)+(5)+(6):
a^4-a2^2-3a^2+a+b^4-b2^2-3b^2+b+c^4-c2^2-3c^2+c=0
a^4+b^4+c^4=a2^2+b2^2+c2^2+3a^2+3b^2+3c^2-(a+b+c)=0.

you should be able to do the rest but is your origninal equation correct?

 

[echelon4]

thanks alot guys, really cleared things up for me. Looks like the direction i was trying to take was completely off....