Question 695486
Let {{{ f(y) = y^3+(m-2)y^2+(m-7)y-4 }}}

By the factor theorem {{{ y-A }}} is a factor of {{{ f(y) }}} if and only if {{{ f(A)=0 }}}

So {{{ y+1 }}} is a if and only if {{{ f(-1) = 0 }}}, Now
{{{ f(-1) = (-1)^3+(m-2)(-1)^2+(m-7)(-1)-4 }}}
{{{ f(-1) =  -1 + (m-2) - (m-7) - 4 }}}
{{{ f(-1) =  0 }}} 
so {{{ f(-1) = 0 }}} showing {{{ y+1 }}}  is a factor QED


Similarly, {{{ f(-2)=0 }}} so,
{{{ (-2)^3 + (m-2)(-2)^2 + (m-7)(-2) - 4 = 0 }}}
{{{ -8 + 4(m-2) -2(m-7) - 4 = 0 }}}
{{{ -8 + 4m-8 -2m +14 - 4 = 0 }}}
{{{ 2m - 6 =0  }}} giving {{{ m = 3 }}}

So substituting {{{ m = 3 }}}, gives:
{{{ f(y) = y^3 + (3-2)y^2 + (3-7)y - 4 }}}
{{{ f(y) = y^3 + y^2 - 4y - 4 }}}

We now know that both {{{ y + 1 }}} and {{{ y + 2 }}} are factors
Suppose the last factor is {{{ y+n }}} then
{{{ f(y) = y^3 + y^2 - 4y - 4 }}}
{{{ f(y) = (y+1)(y+2)(y+n) }}}

Multiplying out gives:
{{{ f(y) = (y^2 + 3y + 2)(y+n) }}}
{{{ f(y) = y^3 + 3y^2 + 2y + ny^2 + 3ny + 2n }}}
{{{ f(y) = y^3 + (3+n)y^2 + (3n+2)y + 2n }}}

So Comparing with the constant coefficient of {{{ f(y) = y^3 + y^2 - 4y - 4 }}}
gives: {{{ 2n = -4 }}} so {{{ n=-2 }}}

Hence the third factor is {{{ y-2 }}}