First, get a common denominator to combine the right side: ‌
A
x+2
+‌
Bx+C
x2+1
=‌
A(x2+1)+(Bx+C)(x+2)
(x+2)(x2+1)
Set the numerators equal to each other: x2−3=A(x2+1)+(Bx+C)(x+2) Expand the right side: ‌A(x2+1)=Ax2+A ‌(Bx+C)(x+2)=Bx2+2Bx+Cx+2C=Bx2+(2B+C)x+2C Combine all terms on the right: ‌Ax2+Bx2=(A+B)x2 ‌(2B+C)x ‌A+2C So we get: x2−3=(A+B)x2+(2B+C)x+(A+2C) Now, match the coefficients for x2,x, and the constant term: For x2:A+B=1 For x:2B+C=0 For constant: A+2C=−3 Write the equations: 1. A+B=1 2. 2B+C=0 3. A+2C=−3 Now solve these equations step by step: From equation 2: 2B+C=0⟶C=−2B Substitute C=−2B into equation 3: A+2(−2B)=−3A−4B=−3 A−4B=−3 Now use A+B=1. Substitute A=1−B into the last equation: ‌1−B−4B=−3 ‌1−5B=−3 ‌−5B=−4 ‌B=‌