How do you find all the zeros of #f(x)=x^4-8x^3+17x^2-8x+16 #?

1 Answer
May 1, 2016

Answer:

Zeros at #x in {4,i,-i}#

Explanation:

Given
#color(white)("XXX")color(red)(1)x^4-8x^3+17x^2-8x+color(blue)(16)#

By the rational root theorem, we know that any rational roots must be factors of #color(blue)(16)/color(red)(1) =16#
which implies they must be in the set #{+-1,+-2,+-4,+-16}#

Testing each of these possibilities give a zero at (an only at) #x=4#

If we divide #x^4-8x^3+17x^2-8x+15# by #(x-4)# (using synthetic or long polynomial division)
we get factors #(x-4)(x^3-4x^2+x-4)#

Again, applying the factor theorem of #x^3-4x^2+x-4)#
we find the only zero at #x=4# (again).

Dividing #x^3-4x^2+x-4# by #(x-4)#
we can re-factor as #(x-4)(x-4)(x^2+1)#

#(x^2+1)# has no Real zeros but can be factored as #(x+i)(x-i)# to get complex zeros at #+-i#