What are all the zeroes of #f(x)= x^3 -4x^2 +6x - 4#?

1 Answer
Mar 10, 2018

Answer:

#f(x)# has zeros #x=2# and #x=1+-i#

Explanation:

Given:

#f(x) = x^3-4x^2+6x-4#

Rational root theorem

By the rational root theorem any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-4# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-4#

Descartes' Rule of Signs

The pattern of signs of the coefficients of #f(x)# is #+ - + -#. With #3# changes of signs, Descartes' Rule of Signs tells us that #f(x)# has #3# or #1# positive real zeros.

The pattern of signs of the coefficients of #f(-x)# is #- - - -#. With no changes of signs, Descartes' Rule of Signs tells us that #f(x)# has no negative real zeros.

So the only possible rational zeros are the positive ones:

#1, 2, 4#

We find:

#f(2) = 8-16+12-4 = 0#

So #x=2# is a zero and #(x-2)# a factor:

#x^3-4x^2+6x-4 = (x-2)(x^2-2x+2)#

The zeros of the remaining quadratic are non-real complex, but we can find them by completing the square:

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

#color(white)(x^2-2x+2) = (x-1)^2-i^2#

#color(white)(x^2-2x+2) = ((x-1)-i)((x-1)+i)#

#color(white)(x^2-2x+2) = (x-1-i)(x-1+i)#

So the remaining zeros are:

#x = 1+-i#