Let us factorise the following polynomials:

x³ – 9x² + 23x – 15

Given, f(x)= x³ – 9x² + 23x – 15

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

f(1)=4.(1)³ −9.(1)² +3.(1)+2=0

We observe that f(1) = 0

From factor theorem, we can say, (x−1) is a factor of f(x)

x³ – 9x² + 23x – 15= x³ – 9x² + 23x – 15

= x³−x²−8x²+8x+15x−15

= x²(x−1)−8x(x−1)+15(x−1)

= (x−1)(x²−8x+15)

= (x−1)(x²−5x−3x+15)

= (x−1)( x(x−5) −3(x−5) )

= (x−1)(x−5)(x−3)