Question 172463
Complex solutions to polynomial always come in conjugate pairs. 
So, if 4-i is a solution, then so is 4+i. 
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I usually graph functions to try get a better sense of where the zeros are.
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{{{ graph( 300, 300, -5, 5, -100, 100, x^4-8x^3+19x^2-16x+34) }}} 
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As you see here, the function does get close to crossing the x axis which means all of the roots are complex. 
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Since we know that (4-i) and (4+i) are both roots, then 
{{{(x-(4+i))(x-(4-i))=x^2-8x+17}}}
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We can divide (using polynomial long division) the original polynomial with this polynomial to find the remainder and then use the quadratic formula on the remainder to find its roots.
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{{{(x^4-8x^3+19x^2-16x+34)/(x^2-8x+17)}}}
First multiplier: {{{highlight(x^2)(x^2-8x+17)=x^4-8x^3+17x^2}}}
Remainder:{{{(x^4-8x^3+19x^2-16x+34)-(x^4-8x^3+17x^2)=2x^2-16x+34}}}
Second multiplier:{{{highlight(2)(x^2-8x+17)=2x^2-16x+34}}}
Remainder:{{{(2x^2-16x+34)-(2x^2-16x+34)=0}}}
The other polynomial is {{{x^2+2}}}
{{{x^4-8x^3+19x^2-16x+34=(x^2-8x+17)(x^2+2)}}}
The roots are:
{{{x^2+2=0}}}
{{{x^2=-2}}}
{{{x=0 +- sqrt(2)i}}}
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So the 4 roots of {{{x^4-8x^3+19x^2-16x+34}}} are
{{{x=4-i}}}
{{{x=4+i}}}
{{{x=sqrt(2)i}}}
{{{x=-sqrt(2)i}}}