is this factorable?

spacewater

Junior Member
Joined
Jul 10, 2009
Messages
67
problem
find the zeros algebraically
f(x)=3x212x+3\displaystyle f(x) = 3x^2-12x+3

steps
f(x)=3(x24x+1)\displaystyle f(x)= 3(x^2-4x+1)

this is farthest i got. How can I factor this problem?
 
spacewater said:
problem
find the zeros algebraically
f(x)=3x212x+3\displaystyle f(x) = 3x^2-12x+3

steps
f(x)=3(x24x+1)\displaystyle f(x)= 3(x^2-4x+1) ? This function has real but irrational roots. That means this function can be factored - but not with "nice" rational numbers.

this is farthest i got. How can I factor this problem?
 
spacewater said:
problem
find the zeros algebraically
f(x)=3x212x+3\displaystyle f(x) = 3x^2-12x+3
steps
f(x)=3(x24x+1)\displaystyle f(x)= 3(x^2-4x+1)
It is not too bad.
3(x44x+1)=3(x23)(x2+3)\displaystyle 3(x^4-4x+1)=3(x-2-\sqrt{3})(x-2+\sqrt{3})
 
spacewater said:
find the zeros algebraically
f(x)=3x212x+3\displaystyle f(x) = 3x^2-12x+3

How can I factor this problem?
Do the instructions specify that you must factor to find the zeroes? Or are you allowed to use the Quadratic Formula instead? :?:
 
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