Answer :

Given, f(a)= 5a^{3} + 11a^{2} + 4a –2

In f(a) putting a=±1, ±2, ±3, we see for which value of a, f(a)=0

f(−1)=5.(−1)^{3}+11.(−1)^{2}+4.(−1)−2=0

We observe that f(−1) = 0

From factor theorem, we can say, (a+1) is a factor of f(a)

5a^{3} + 11a^{2} + 4a –2 = 5a^{3} + 11a^{2} + 4a –2

= 5a^{3}+5a^{2}+6a^{2}+6a−2a−2

= 5a^{2}(a+1)+6a(a+1)−2(a+1)

= (a+1)(5a^{2}+6a−2)

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