Q. polynomial-roots, etc (1 Viewer)

[marsenal]

The equation x³ + 2x + 1=0 has roots a, b, and c. Evaluate a⁴ + b⁴ + c⁴

So we know that,
a + b + c= 0
ab + ac + bc=2
abc= -1

But using these I am still not able to get the final solution. I get:

a⁴ + b⁴ + c⁴ = (a+b+c)⁴ - 2a³(b+c) - 2b³(a+c) - 2c(a+b) - 5(a²b² + a²c² + b²c²) - 6abc (a +b +c)

From here I just keep stuffing it up ( I am assuming that so far I have got everything right).

 

[spice girl]

Technique1:

Find the polynomial with roots a^2, b^2 and c^2

Then you can find (a^2)^2 + (b^2)^2 + (c^2)^2 using the normal way of finding the sum-of-squares method.

Technique 2:
Be an absolute freak and remember by heart that:
a^4 + b^4 + c^4 = (a^3 + b^3 + c^3)(a+b+c) - (a^2 + b^2 + c^2)(ab + bc + ac) + abc(a + b + c)

 

[bobo123]

erm
heres an alternative
its longer
but yeah im not on spice girls level


sub a,b,c into p(x) individually

so you get
a^3+2a+1 = 0 (1)
b^3+2b+1 = 0 (2)
c^3+2c+1 = 0 (3)

mutiply (1) by a, (2) by b and (3) by c
so you get
a^4+2a^2+a = 0
b^4+2b^2+b = 0
c^4+2c^2+c = 0

all the lot together

you get :

(a^4+b^4+c^4) = -2(a^2+b^2+c^2) - (a+b+c)

answer is -7?

 

[ezzy85]

Originally posted by bobo123
erm
heres an alternative
its longer
but yeah im not on spice girls level


sub a,b,c into p(x) individually

so you get
a^3+2a+1 = 0 (1)
b^3+2b+1 = 0 (2)
c^3+2c+1 = 0 (3)

mutiply (1) by a, (2) by b and (3) by c
so you get
a^4+2a^2+a = 0
b^4+2b^2+b = 0
c^4+2c^2+c = 0

all the lot together

you get :

(a^4+b^4+c^4) = -2(a^2+b^2+c^2) - (a+b+c)

answer is -7?

using this method i got a⁴ + b⁴ + c⁴ = 8

 

[bobo123]

yeah sorry my mistake