Question 474274
{{{ 6x^3 - 16x^2 - 6x }}}
{{{ 2x*(3x^2 - 8x - 3) }}}
Find the factors of the quadratic
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{ a = 3 }}}
{{{ b = -8 }}}
{{{ c = -3 }}}
{{{x = (-(-8) +- sqrt( (-8)^2-4*3*(-3) ))/(2*3) }}} 
{{{x = ( 8 +- sqrt( 64 + 36 )) / 6 }}} 
{{{x = ( 8 +- sqrt( 100 )) / 6 }}} 
{{{x = ( 8 + 10) / 6 }}} 
{{{ x = 18/6 }}}
(1) {{{ x = 3 }}}
and
{{{x = ( 8 - 10) / 6 }}} 
{{{ x = -2/6 }}}
(2) {{{ x = -1/3 }}}
Now rewrite (1) and (2) as:
(1) {{{ x - 3 = 0 }}}
(2) {{{ x + 1/3 = 0 }}}
(2) {{{ 3x + 1 = 0 }}}
so,
{{{ 3x^2 - 8x - 3  = ( x - 3 )*( 3x + 1 ) }}}
The complete factorization is:
{{{ 2x*( x - 3 )*( 3x + 1 ) }}}
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check:
{{{ 2x*(3x^2 - 9x + x - 3) }}}
{{{ 2x*(3x^2 - 8x - 3) }}}
{{{ 6x^3 - 16x^2 - 6x }}}
OK