x2+3x+2=(x+2)(x+1))
Since f(x) is divisible by (x+2)(x+1)
f(−2)&(f−1) must be zero \rightarrow Remainder theorem
f(−2)=(−2)3+a(−2)2−b(−2)+10=0−8+4a+2b+10=0
or 4a+2b+2=0 -----(1)
f(−1)=(−1)3+a(−1)2−b(−1)+10=0=−1+a+b+10=0a+b+9=0 -----(2)
Solving (1)&(2)
put a=−(b+a)is(1)−4(b+9)+2b+2=0−4b−36+2b+2=0−2b−34=0orb=17a=−(17+9)=−26