Question 589115
{{{ x = -2i }}}
{{{ x = 2i }}}
{{{ x = 3 }}}
{{{ f(1) = 20 }}}
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{{{ ( x - 2i )*( x + 2i )*( x - 3 ) = 0
{{{ ( x^2 - 4*i^2)*( x - 3 ) = 0 }}}
Note that {{{ i^2 = -1 }}}
{{{ ( x^2 + 4 )*( x - 3 ) = 0 }}}
{{{ x^3 + 4x - 3x^2 - 12 = 0 }}}
{{{ k*( x^3 - 3x^2 + 4x - 12 ) = k*0 }}}
{{{ y =  k*( x^3 - 3x^2 + 4x - 12 ) }}}
{{{ y(1) = k*( 1 - 3 + 4 - 12 ) }}}
{{{ y(1) =-10k }}}
{{{ -10k = 20 }}}
{{{ k = -2 }}}
{{{ y =  -2*( x^3 - 3x^2 + 4x - 12 ) }}}
{{{ y = -2x^3 + 6x^2 -8x + 24 }}}
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I think of imaginaries as a way to get off
the real number line by rotating 90 degrees
each time you multiply a real by {{{ i }}}
So, if you multiply twice, that puts you
back on the real number line going
180 degrees the other way, or
multiplying by {{{ -1 }}}