Q. 34.7( 17 Votes )

# Let us factorise the following polynomials:

a^{3} – 12a – 16

Answer :

Given, f(a) = a^{3} – 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)

a^{3} – 12a – 16 = a^{3} – 12a – 16

= a^{3}+2a^{2}–2a^{2}– 4a – 8a − 16

= a^{2}(a+2)−2a(a+2)−8(a+2)

= (a+2)(a^{2}−2a−8)

= (a+2)(a^{2}−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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