Question 135973
Actually, technically speaking, and we should always speak technically when talking about this stuff, {{{z^4 - 8z^3 + 28z^2 - 48z + 35}}} doesn't have any roots.  {{{z^4 - 8z^3 + 28z^2 - 48z + 35 = red(0)}}} has roots, and those roots are related to the factors of {{{z^4 - 8z^3 + 28z^2 - 48z + 35}}} as we shall see shortly.


Assume that {{{2-i}}} is a root of {{{z^4 - 8z^3 + 28z^2 - 48z + 35=0}}}


Then {{{2+i}}} must also be a root of {{{z^4 - 8z^3 + 28z^2 - 48z + 35=0}}} because complex roots always occur in conjugate pairs.


{{{a}}} is a root of a polynomial <i>equation</i> if and only if {{{x-a}}} is a factor of the polynomial.  That means that both {{{z-(2-i)}}} and {{{z-(2+i)}}} are factors of {{{z^4 - 8z^3 + 28z^2 - 48z + 35}}}.  Furthermore, the product of those two factors must also be a factor of {{{z^4 - 8z^3 + 28z^2 - 48z + 35}}}.


The product of {{{z-(2-i)}}} and {{{z-(2+i)}}}, {{{(z-(2-i))(z-(2+i))=(z-2+i)(z-2-i)}}} works out to {{{z^2-2z-zi-2z+4+2i+zi-2i-i^2=z^2-4z+5}}} (recalling that {{{i^2=-1}}} so {{{-(i^2)=-(-1)=+1}}})


Since {{{z^2-4z+5}}} is a factor of {{{z^4 - 8z^3 + 28z^2 - 48z + 35}}}, the result of polynomial long division of {{{z^4 - 8z^3 + 28z^2 - 48z + 35}}} by {{{z^2-4z+5}}} will be a quotient polynomial with a zero remainder.  If this is true, then our original assumption is true, i.e., {{{2-i}}} is a root of {{{z^4 - 8z^3 + 28z^2 - 48z + 35=0}}} because we will have proven that {{{z-(2-i)}}} is a factor of {{{z^4 - 8z^3 + 28z^2 - 48z + 35}}}


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Polynomial Long Division Process
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{{{z^2}}} goes into {{{z^4}}} {{{z^2}}} times.  {{{green(z^2)}}} is the first term of the quotient.


Multiply {{{z^2}}} times {{{z^2-4z+5}}} to get {{{z^4-4z^3+5z^2}}} and subtract from {{{z^4 - 8z^3 + 28z^2}}} resulting in {{{-4z^3+23z^2}}}.  Bring down the {{{-48z}}} to form {{{-4z^3+23z^2-48z}}}


{{{z^2}}} goes into {{{-4z^3}}} {{{-4z}}} times.  {{{green(-4z)}}} is the second term of the quotient.


Multiply {{{-4z}}} times {{{z^2-4z+5}}} to get {{{-4z^3+16z^2-20z}}} and subtract from {{{-4z^3+23z^2-48z}}} resulting in {{{7z^2-28z}}}.  Bring down the {{{35}}} to form {{{7z^2-28z+35}}}


{{{z^2}}} goes into {{{7z^2}}} {{{7}}} times.  {{{green(7)}}} is the third term of the quotient.


Multiply {{{7}}} times {{{z^2-4z+5}}} to get {{{7z^2-28z+35}}} and subtract from {{{7z^2-28z+35}}} resulting in a remainder of zero.  The quotient is the sum of the three parts noted above, namely {{{green(z^2-4z+7)}}}


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End of Polynomial Long Division Process
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That means that {{{z^2-4z+5}}} is a factor of {{{z^4 - 8z^3 + 28z^2 - 48z + 35}}}, and therefore {{{z-(2-i)}}} must also be a factor of {{{z^4 - 8z^3 + 28z^2 - 48z + 35}}}, and finally {{{z=2-i}}} (and its conjugate, {{{z=2+i}}})must be a root of {{{z^4 - 8z^3 + 28z^2 - 48z + 35=0}}}.


Since {{{z^4 - 8z^3 + 28z^2 - 48z + 35=0}}} is a quartic (degree 4) polynomial equation, there must be 4 roots.  We have found two of them.  Since the quotient derived by the polynomial long division process just performed must also be a factor of {{{z^4 - 8z^3 + 28z^2 - 48z + 35}}}, the factors of that quotient must also be factors of {{{z^4 - 8z^3 + 28z^2 - 48z + 35}}}.


Therefore solving {{{z^2-4z+7=0}}} will give the remaining two roots of {{{z^4 - 8z^3 + 28z^2 - 48z + 35=0}}}.  I'll leave that one to you.  Hint: The quadratic does NOT factor over the integers (or the reals, for that matter), so either complete the square or use the quadratic formula.  I completed the square and that worked out rather tidily.  Write back and tell me your results.


The following graph of {{{f(x)=x^4 - 8x^3 + 28x^2 - 48x + 35}}} supports the notion that there are no real roots to your given equation because the graph does not intersect the x-axis, so {{{f(x)<>0}}} for all real x.



{{{
drawing(600,600,-1,9,-1,9,
grid(1),
graph(600,600,-1,9,-1,9,x^4 - 8x^3 + 28x^2 - 48x + 35)
)
}}}