Q. 84.6( 12 Votes )

Let us factorise

Answer :

Given, f(a)= 5a3 + 11a2 + 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)


5a3 + 11a2 + 4a –2 = 5a3 + 11a2 + 4a –2


= 5a3+5a2+6a2+6a−2a−2


= 5a2(a+1)+6a(a+1)−2(a+1)


= (a+1)(5a2+6a−2)


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