Let us factorise the following polynomials:

x³ – 19x – 30

Given, f(x)= x³ – 19x – 30

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

f(−2)=(−2)³−19.(−2)+30=0

We observe that f(−2) = 0

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

x³ – 19x – 30 = x³ – 19x – 30

= x³+2x²−2x²−4x−15x −30

= x²(x+2)−2x(x+2)−15(x+2)

= (x+2)(x²−2x−15)

= (x+2)(x²−5x+3x−15)

= (x+2)( x(x−5) + 3(x−5) )

= (x+2)(x−5)(x+3)