Question 475872
given:
{{{ x = -1 }}}
{{{ x + 1 = 0 }}}
and
{{{ x = 3 }}}
{{{ x - 3 = 0 }}}
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Complex roots come in pairs
of the form {{{ a + bi }}}
and {{{ a - bi }}}
I can factor out {{{ (x + 1)*(x - 3) }}}
{{{ (x + 1)*(x - 3) = x^2 - 2x - 3 }}}
{{{ (x^4 - 2x^2 - 16x - 15) / (x^2 - 2x - 3) = x^2 + 2x + 5 }}}
I can use quadratic formula now
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ a = 1 }}}
{{{ b = 2 }}}
{{{ c = 5 }}}
{{{x = (-2 +- sqrt( 2^2-4*1*5 ))/(2*1) }}}
{{{x = (-2 +- sqrt( 4 - 20 ))/2 }}}
{{{x = (-2 +- sqrt( -16 ))/2 }}}
{{{x = (-2 + 4i)/2 }}}
{{{ x = -1 + 2i }}} 1st root 
and
{{{x = (-2 - 4i)/2 }}}
{{{ x = -1 - 2i }}} 2nd root