Question 825879
When I do the division {{{ ( x^3 - kx^2 + 3x + 7k  ) / ( x + 2 ) }}}, 
I get {{{ x^2 - ( k+2 )*x + 2k + 7 }}} with a remainder of
{{{ ( 3k - 14 )/( x+2 ) }}} I want to make the remainder zero, so
{{{ 3k - 14 = 0 }}}
{{{ 3k = 14 }}}
{{{ k = 14/3 }}}
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check:
{{{ (x + 2 )*( x^2 - ( k+2 )*x + 2k + 7 ) }}} 
{{{ (x + 2 )*( x^2 - ( 14/3+2 )*x + 2*(14/3) + 7 ) }}} 
{{{ (x + 2 )*( x^2 - ( 20/3 )*x + 28/3 + 21/3 ) }}} 
{{{ (x + 2 )*( x^2 - ( 20/3 )*x + 49/3 ) }}} 
This should equal the original equation with {{{ k = 14/3 }}}- you can finish