Answer :
Formula used.
If f(x) is a polynomial with degree n
Then (x – a) is a factor of f(x) if f(a) = 0
1st we find out zero of both polynomial (x + 1)(x + 2)
x + 1 = 0 and x + 2 = 0
x = – 1 and x = – 2
if x + 2 is factor of f(x) = x3 + 3x2 + 2ax + b
then f( – 2) = 0 ;
f( – 2) = ( – 2)3 + 3( – 2)2 + 2a( – 2) + b
= – 8 + 12 – 4a + b
b – 4a + 4 = 0
b = 4a – 4 ………eq 1
if x + 1 is factor of f(x) = x3 + 3x2 + 2ax + b
then f( – 1) = 0 ;
f( – 1) = ( – 1)3 + 3( – 1)2 + 2( – 1)a + b
= – 1 + 3 – 2a + b
b – 2a + 2 = 0 ………eq 2
Putting value of b from eq 1 into eq 2
(4a – 4) – 2a + 2 = 0
2a – 2 = 0
2a = 2
a = = 1
Putting value of a in eq 1
b = 4 × (1) – 4
= 0
Conclusion.
∴ if (x + 1) and (x + 2) is factor of f(x) then b = 0 and a = 1
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