Question 175322
{{{(y-5)/(2y^2 - 7y - 15)}}}
Factor the polynomial {{{(2y^2 - 7y - 15)}}}
{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{a = 2}}}
{{{b = -7}}}
{{{c = -15}}}
{{{y = (-(-7) +- sqrt((-7)^2-4*2*(-15) ))/(2*2) }}}
{{{y = (7 +- sqrt(49 + 120 ))/4 }}}
{{{y = (7 +- 13)/4 }}}
{{{y = 5}}}
and
{{{y = -(3/2)}}}
The factors are:
{{{(y - 5)(y + 3/2) = 2y^2 - 7y - 15}}}
This is the same as
{{{(y - 5)(2y + 3) = 2y^2 - 7y - 15}}}
and so,
{{{(y - 5)/(2y^2 - 7y - 15) = (y - 5) / (y - 5)(2y + 3) }}}
{{{(y - 5) / (y - 5)(2y + 3) = 1 / (2y + 3)}}} answer