Let A and B be real numbers such that $\frac{A}{x-5}+B(x+1)=\frac{-3x^2+12x+22}{x-5}$. What is A+B?
+600 Note that by synthetic division $-3 x^2 + 12 x + 22 = (-3 x - 3)(x - 5) + 7 = -3(x+1)(x-5)+7$. Divide both sides by $x-5$.
+26367 Let A and B be real numbers such that
\(\dfrac{A}{x-5}+B*(x+1)=\dfrac{-3x^2+12x+22}{x-5}\).
What is A+B?
\(\small{ \begin{array}{|lrcll|} \hline & \dfrac{A}{x-5}+B*(x+1) &=&\dfrac{-3x^2+12x+22}{x-5} \quad | \quad *(x-5) \\\\ & \mathbf{ A+B*(x+1)(x-5) } &=& \mathbf{ -3x^2+12x+22 } \\ \hline 1.~ x=-1:& A + 0 &=& -3*(-1)^2 + 12*(-1) + 22 \\\\ & A &=& -3-12+22 \\ & \mathbf{A} &=& \mathbf{ 7 } \\ \hline 2.~ x =-2:& 7+B*(-1)(-7) &=& -3*(-2)^2 + 12*(-2) + 22 \\ & 7+7B &=& -12-24+22 \\ & 7B &=& -14-7 \\ & 7B &=& -21 \quad | \quad : 7 \\ & \mathbf{B} &=& \mathbf{ -3 } \\ \hline & A+B &=& 7-3 \\ & \mathbf{ A+B } &=& \mathbf{ 4 } \\ \hline \end{array} }\)
