The roots of x^3+6x^2+4x+2=0 are r, s, and t. For what order triple of integers (a,b,c) are 2r, 2s, and 2t the roots of x^3+ax^2+bx+c=0?

Question

roots of x^3+6x^2+4x+2=0 doubled equals the coefficients in x^3+ax^2+bx+c=0

Answer 1

r^3+6r^2 + 4r + 2 = 0

 

s^3 +6s^2 + 4s + 2 = 0

 

t^2 + 6t^2 + 4t + 2 = 0

 

 

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(2r)^3 + a(2r)^2 + b(2r) + c = 0

8r^3 + (4a)r^2 + (2b)r + c = 0

4r^3 + (2a)r^2 + (b)r + c = 0 <---- divides everything by 2

r^3 + ((1/2)a) r^2 + (b/4)r + (c/4) = 0 <---- divides everything by 4

r^3 +     6       r^2 +    4  r  +   2 = 0    <---- equates the coefficient to the given equation

 

(1/2)A = 6 ---> A=12

b/4 = 4  ---> B = 16

c/4 = 2 ----> C = 8