Question 317215
{{{ graph( 300, 300,-5,5,-8,8, x^7-x^5-16x^3+16x) }}} 
As you see from the graph, there are 5 real roots.
x=-2,-1,0,1,2
Each root has multiplicity of 1.
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The polynomial found by these zeros is,
{{{(x+2)(x+1)x(x-1)(x-2)=x(x^2-1)(x^2-4)=x(x^4-4x^2-x^2+4)=x^5-5x^3+4x}}}
Divide the original polynomial by this polynomial to find the quadratic remainder.
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First factor:{{{x^2}}}
{{{x^2(x^5-5x^3+4x)=x^7-5x^5+4x^3}}}
Subtract this from the original polynomial to get the remainder,
{{{(x^7-x^5-16x^3+16x)-(x^7-5x^5+4x^3)=4x^5-20x^3+16x}}}
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Next factor:{{{4}}}
{{{4(x^5-5x^3+4x)=4x^5-20x^3+16x}}}
Subtract this from the remainder
{{{(4x^5-20x^3+16x)-(4x^5-20x^3+16x)=0}}}
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So the remainder quadratic is {{{x^2+4=0}}} which has complex roots ({{{-2i}}},{{{2i}}})
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{{{x^7-x^5-16x^3+16x=(x+2)(x+1)x(x-1)(x-2)(x^2+4)}}}