Question 472123
{{{ 3x^2 + 25x + 50 }}}
The direct way is to use the quadratic formula to find the roots
The formula is:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
when the form of the equation is:
{{{ ax^2 + bx + c = 0 }}}
{{{ a = 3 }}}
{{{ b = 25 }}}
{{{ c = 50 }}}
{{{x = (-25 +- sqrt( 25^2-4*3*50 ))/(2*3) }}} 
{{{x = (-25 +- sqrt( 625 - 600 )) / 6 }}} 
{{{x = (-25 +- sqrt( 25 )) / 6 }}} 
{{{ x = ( -25 + 5 )/6 }}}
{{{ x = -10 / 3 }}}
and
{{{ x = ( -25 - 5) / 6 }}}
{{{ x = -5 }}}
The roots are {{{-10/3}}} and {{{ -5 }}}
I can rewrite the solutions as:
{{{ x + 10/3 = 0 }}} and
{{{ x + 5 = 0 }}}
And If I multiply these together:
{{{ (x + 10/3)*(x + 5) = 0 }}}
Expanding:
{{{x^2 + (10/3)*x + 5x + 50/3 = 0 }}}
Multiply both sides by {{{ 3 }}}
{{{3x^2 + 10x + 15x + 50 = 0 }}}
{{{ 3x^2 + 25x + 50 = 0 }}}
So the factors of the equation are:
{{{ x + 10/3  }}} and
{{{ x + 5  }}}