How do you find all the zeros of #g(x) = x^5 – 8x^4 + 28x^3 – 56x^2 + 64x – 32 #?

#g(x)=x^5-8x^4+28x^3-56x^2+64x-32#

By the rational root theorem, any rational zeros of #g(x)# must be expressible in the form #p/q# for integers #p# and #q# with #p# a divisor of the constant term #32# and #q# a divisor of the coefficient #1# of the leading term.

In addition note that #g(-x) = -x^5-8x^4-28x^3-56x^2-64x-32# has no changes of sign in its coefficients, so there are no negative Real zeros.

Hence the only possible rational zeros are:

#1, 2, 4, 8, 16, 32#

We find:

#g(2) = 32-128+224-224+128-32 = 0#

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

#x^5-8x^4+28x^3-56x^2+64x-32#

#=(x-2)(x^4-6x^3+16x^2-24x+16)#

Let #h(x) = x^4-6x^3+16x^2-24x+16#

#h(2) = 16-48+64-48+16 = 0#

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

#x^4-6x^3+16x^2-24x+16 = (x-2)(x^3-4x^2+8x-8)#

Let #k(x) = x^3-4x^2+8x-8#

#k(2) = 8-16+16-8 = 0#

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

#x^3-4x^2+8x-8 = (x-2)(x^2-2x+4)#

The remaining quadratic factor can be factored by completing the square and using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a = x - 1# and #b=sqrt(3)i#

#x^2-2x+4#

#= x^2-2x+1+3#

#= (x-1)^2-(sqrt(3)i)^2#

#= ((x-1)-sqrt(3)i)((x-1)+sqrt(3)i)#

#= (x-1-sqrt(3)i)(x-1+sqrt(3)i)#

Hence the last two zeros are:

#x = 1+-sqrt(3)i#