What are all the possible rational zeros for #f(x)=3x^3-7x^2+29x-9# and how do you find all zeros?
1 Answer
The "possible" rational zeros are:
#+-1/3, +-1, +-3, +-9#
The actual zeros are:
#1/3# and#1+-2sqrt(2)i#
Explanation:
#f(x) = 3x^3-7x^2+29x-9#
By the rational roots theorem, any rational zeros of
That means that the only possible rational zeros are:
#+-1/3, +-1, +-3, +-9#
Note also that the signs of the coefficients of
So the only possible rational real zeros are:
#1/3, 1, 3, 9#
Trying each in turn we immediately find:
#f(1/3) = 3(1/27)-7(1/9)+29(1/3)-9 = (1-7+87-81)/9 = 0#
So
#3x^3-7x^2+29x-9 = (3x-1)(x^2-2x+9)#
We can factor the remaining quadratic by completing the square, but it does require some Complex coefficients:
#x^2-2x+9 = x^2-2x+1+8#
#color(white)(x^2-9x+9) = (x-1)^2-(2sqrt(2)i)^2#
#color(white)(x^2-9x+9) = ((x-1)-2sqrt(2)i)((x-1)+2sqrt(2)i)#
#color(white)(x^2-9x+9) = (x-1-2sqrt(2)i)(x-1+2sqrt(2)i)#
Hence the other two zeros are:
#x = 1+-2sqrt(2)i#