(x^2-x-8)/((x+1)(x^2+5x+6))x2−x−8(x+1)(x2+5x+6)
First we need to finish factoring the denominator:
(x^2-x-8)/((x+1)(x+2)(x+3))x2−x−8(x+1)(x+2)(x+3)
Now we want A, B, "and " CA,B,and C to get:
A/(x+1) + B/(x+2) + C/(x+3) = (x^2-x-8)/((x+1)(x+2)(x+3))Ax+1+Bx+2+Cx+3=x2−x−8(x+1)(x+2)(x+3)
So we need:
A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2) = x^2-x-8A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2)=x2−x−8
The left side can be rewritten:
Ax^2+5Ax+6A+Bx^2+4Bx+3B+Cx^2+3Cx+2CAx2+5Ax+6A+Bx2+4Bx+3B+Cx2+3Cx+2C
and then:
(A+B+C)x^2 +(5A+4B+3C)x+(6A+3B+2C)=x^2-x-8(A+B+C)x2+(5A+4B+3C)x+(6A+3B+2C)=x2−x−8
So we need to solve the system:
A+B+C=1A+B+C=1
5A+4B+3C=-15A+4B+3C=−1
6A+3B+2C=-86A+3B+2C=−8
Multiplying the first equation by -3−3 and adding to the second. Then the first times -2−2 and add to the third, gets us:
2A+B=-42A+B=−4
4A+B=-104A+B=−10
So 2A = -62A=−6 and A = -3A=−3, which gets us B = 2B=2 and together these give us C=2C=2.
(x^2-x-8)/((x+1)(x+2)(x+3))= -3/(x+1) + 2/(x+2) + 2/(x+3) x2−x−8(x+1)(x+2)(x+3)=−3x+1+2x+2+2x+3
