Q. 34.7( 17 Votes )

Let us factorise the following polynomials:

a3 – 12a – 16

Answer :

Given, f(a) = a3 – 12a – 16

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


f(−2)=(−2)3−12.(−2)−16=0


We observe that f(−2) = 0


From factor theorem, we can say, (a+2) is a factor of f(a)


a3 – 12a – 16 = a3 – 12a – 16


= a3+2a2–2a2– 4a – 8a − 16


= a2(a+2)−2a(a+2)−8(a+2)


= (a+2)(a2−2a−8)


= (a+2)(a2−4a+2a−8)


= (a+2)( a(a−4) + 2(a−4) )


= (a+2)(a−4)(a+2)


= (a+2)2(a−4)


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