How do you find all the zeros of #h(x) = x³ - 3x² + 4x - 2#?

1 Answer
Jun 5, 2016

Answer:

#x=1# or #x = 1+-i#

Explanation:

#h(x) = x^3-3x^2+4x-2#

Note that the sum of the coefficients is #0#. That is:

#1-3+4-2 = 0#

So #h(1) = 0#, #x=1# is a zero and #(x-1)# is a factor:

#x^3-3x^2+4x-2 = (x-1)(x^2-2x+2)#

The remaining quadratic has negative discriminant, so only Complex zeros, but we can complete the square and use the difference of squares identity to find them:

#x^2-2x+2#

#= x^2-2x+1+1#

#= (x-1)^2-i^2#

#= ((x-1)-i)((x-1)+i)#

#= (x-1-i)(x-1+i)#

So the other two zeros are #x = 1+-i#