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

1^{st} 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) = x^{3} + 3x^{2} + 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) = x^{3} + 3x^{2} + 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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