Question 360223
since the polynomial is of degree 4, there are 4 zeros or roots.
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By descartes sign rule
f(+x) +-++-  there are 3 or 1 positive real roots
f(-x) +++--  there is exactly 1 negative real root
since there is one negative real root, lets find it
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looking at the constant 10, its factors are -+1,-+2,-+5,-+10 plus all the ones created by dividing these factors by -+1, -+2 from the leading constant 2
so the total possible reals are -+1/2,-+1,-+2,-+5/2,-+5,-+10
usually the roots will not be in the extreme so lest try -1.
f(-1)={{{2(-1)^4-(-1)^3+2*(-1)^2+19(-1)-10=2+1+2-19-10=-24}}} so -1 no a root
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try -2
f(-2)={{{2(-2)^4-(-2)^3+2(-2)^2+19(-2)-10=32+8+8-38-10=0}}} so -2 is a root
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Perform synthetic division to reduce the polynomial
yielding   {{{(x+2)*(2x^3-5x^2+12x-5)}}}
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try another "potential root" from the list, now realizing that there is at least one positive real
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try 1/2
through synthetic division you will find this works.  
so now the functions is factored to  {{{(x+2)*(x-(1/2))*2(x^2-2x+5)}}}
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using the quadratic equation on {{{x^2-2x+5}}} we find that 1+2i and 1-2i are roots or zeros
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So the four roots are -2, 1/2, 1+2i, 1-2i