⇒−x2+6x+13=A(x+2)2+B(3x+5)(x+2)+C(3x+5) ...(i) It is an identity ∴ True for every value of x. Put x=−2 in Eq. (i), −(−2)2+6(−2)+13=A.0+B.0+C(3(−2)+5) ⇒−4−12+13=−C ⇒C=3 Put x=
−5
3
in Eq. (i), we get −(
−5
3
)2+6(
−5
3
)+13 =A(
−5
2
+2)2+B.0+C.0 ⇒
−25
9
−10+13=A.
1
9
−25
9
+3=
A
9
⇒A=2 Put x=0 in Eq. (i), we get 13=4A+10B+5C ⇒13=8+10B+15 ⇒10B=13−23 or B=−1 ∴