Let us factorise the following polynomials:

x³ – 6x + 4

Given, f(x) = x³ – 6x + 4

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

f(2)=(2)³−6.2+4=0

We observe that f(2) = 0

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

x³ – 6x + 4 = x³ – 6x + 4

= x³ −2x² +2x² −4x −2x +4

= x²(x−2)+2x(x−2)−2(x−2)

= (x−2)(x²+2x−2)