# \mbox{We can use The Remainder Theorem with this.} #
# \mbox{Let:} \qquad R = \mbox{the remainder when} \quad
x^2 + (m-2) x - m^2 - 3 m \quad \mbox{ is divided by} \quad x+m. #
# \mbox{Note:} \quad x+m = x - (-m). #
# \mbox{Applying The Remainder Theorem to this:} #
# R = \mbox{given polynomial, with} \ \ x \ \ \mbox{replaced by} \ \ -m. #
# :. \qquad R = (-m)^2 + (m-2)(-m) - m^2 - 3 m #
# :. \qquad R = \color{red}{cancel{m^2}} - \color{red}{cancel{m^2}} + 2 m - m^2 - 3 m #
# :. \qquad R = 2 m - m^2 - 3 m #
# :. \qquad R = -m^2 - m. #
# \mbox{As per the problem, we want:} \qquad R = -1. #
# \mbox{So, equating, we get:} \qquad -m^2 - m = -1. #
# :. \qquad -m^2 - m + 1 = 0. #
# :. \qquad -( -m^2 - m + 1 ) = -( 0 ). #
# :. \qquad m^2 + m - 1 = 0. #
# \mbox{By the Quadratic Formula, we get:} #
# \qquad \qquad qquad \qquad m =
{ -1 \pm \sqrt{ 1^2 - 4 \cdot 1 \cdot (-1) } } / { 2 \cdot 1 } #
# \qquad \qquad qquad \qquad m =
{ -1 \pm \sqrt{ 1 + 4 } } / { 2 }=
{ -1 \pm \sqrt{ 5 } } / { 2 }. #
# \ #
# \mbox{So we have our answers:} #
# \qquad \qquad qquad \qquad \qquad \qquad qquad \qquad m =
{ -1 \pm \sqrt{ 5 } } / { 2 }. #
# \qquad \qquad qquad \qquad \qquad \quad \ \ m =
{ -1 - \sqrt{ 5 } } / { 2 }, \quad { -1 + \sqrt{ 5 } } / { 2 }. #
