Question 320799
From the rational roots theorem, 
{{{p=0 +- 1}}}
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{{{x=1}}}
{{{f(1)=1+3+3+1=8}}}
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{{{x=-1}}}
{{{f(-1)=-1+3-3+1=0}}}
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So {{{x=-1}}} is a root and {{{x+1}}} is a factor of the polynomial.
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Now use polynomial long division to find the remaining quadratic,
First term: {{{highlight( x^2)}}}
{{{x^2(x+1)=x^3+x^2}}}
Subtract that product from the original polynomial to get the remainder,
{{{(x^3+3x^2+3x+1)-(x^3+x^2)=2x^2+3x+1}}}
Second term:{{{highlight(2x)}}}
{{{2x(x+1)=2x^2+2x}}}
Subtract that product from the previous remainder to get the new remainder,
{{{(2x^2+3x+1)-(2x^2+2x)=x+1}}}
Final term:{{{highlight(1)}}}
{{{1(x+1)=x+1}}}
Subtract that product from the previous remainder to get the new remainder,
{{{(x+1)-(x+1)=0}}}
Gather all of the terms.
{{{(x^3+3x^2+3x+1)/(x+1)=x^2+2x+1=(x+1)^2}}}
{{{x^3+3x^2+3x+1=(x+1)^3}}}
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A triple root at {{{x=-1}}}