Identify all zeros of the polynomial function f(x)=x63-3x^2+4x-12

Question

anonymous · Answer

* x^3

anonymous · Answer

not x63

campbell_st · Answer

ok... looks like x = 3 is a zero.. 

f(3) = 0 when you substitute 

so group in pairs

$$f(x) = x^2 + 4x - 3x^2 - 12$$

which becomes

$$f(x) = x(x^2 +4) -3(x^2 + 4)$$

which factors to 

$$f(x) = (x -3)(x^2 + 4)$$

campbell_st · Answer

hope this makes sense...

campbell_st · Answer

oops should read 

$$f(x) = x^3 + 4x - 3x^2 - 12$$

anonymous · Answer

Im not sure what the answer should look like
I got -1,2,-2

campbell_st · Answer

well here is my point, if a polynomial has a zero at x = -1 

then f(-1) = 0

so looking at your solutions

f(-1) = (-1)^3 - 3(-1)^2 + 4(-1) - 12

f(-1) = -1 -3 - 4 -12

f(-1) = -20

x = -1 can't be a zero

f(-2) = (-2)^3 - 3(-2)^2 + 4(-2) - 12

        = -8 -12 - 8 - 12
        = -60 

so x = -2 can't be a zero

f(2) = (2)^3 - 3(2)^2 + 4(2) - 12

      = 8 - 12 + 8 - 12

f(2) = -8

so x = 2 isn't a zero

so perhaps you are looking at the wrong answer.

anonymous · Answer

ok-3 and -4?

campbell_st · Answer

here is a graph of your function

campbell_st · Answer

and - 4 won't work either

$$f(x) = (x -3)(x^2 + 4)$$

the zeros occur when 

x - 3 = 0..... and/or      

$$x^2 + 4 = 0$$

so x = 3 is easy to find as a zero

by now you have the problem of

$$x^2 = -4$$

you are now dealing with complex numbers

as the square root of -4 is undefined in the real number system

campbell_st · Answer

so you function only has 1 real root, x = 3

campbell_st · Answer

hope this all makes sense.

anonymous · Answer

I get it, i just did it all over again thank you so much!