Answer:
3x^2 -2x + 1 =3(x^2-2/3x+1/3)=3(x-1/3)^2+2/9*3= 3(x-1/3)^2+2/3
(x-1/3)^2 is greater or equal to zero
3(x-1/3)^2 is greater or equal to zero
and 2/3 is greater than zero
So there sum is greater than zero
Proved
Step-by-step explanation:
3x^2 -2x + 1 =3(x^2-2/3x+1/3)
Consider x^2-2/3x+1/3
Remember that (a-b)^2 =a^2-2ab+b^2
x^2=a^2
a=x
-2/3x= -2*x*b
b=1/3
S0 (x-1/3)^2= x^2-2/3x+1/9
x^2-2/3x+1/3= x^2-2/3x+1/9+1/3-1/9= (x-1/3)^2+2/9
3x^2 -2x + 1 =3(x^2-2/3x+1/3)=3(x-1/3)^2+2/9*3= 3(x-1/3)^2+2/3
(x-1/3)^2 is greater or equal to zero
3(x-1/3)^2 is greater or equal to zero
and 2/3 is greater than zero
So there sum is greater than zero
Proved
