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?
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 $x-5$.
Let A and B be real numbers such thatAx−5+B∗(x+1)=−3x2+12x+22x−5.What is A+B?
Ax−5+B∗(x+1)=−3x2+12x+22x−5|∗(x−5)A+B∗(x+1)(x−5)=−3x2+12x+221. x=−1:A+0=−3∗(−1)2+12∗(−1)+22A=−3−12+22A=72. x=−2:7+B∗(−1)(−7)=−3∗(−2)2+12∗(−2)+227+7B=−12−24+227B=−14−77B=−21|:7B=−3A+B=7−3A+B=4