Using factor theorem, factorize of the polynomials:

Solution:

Given, $f(x)=3 x^{3}-x^{2}-3 x+1$

The factors of constant term 1 is ± 1

The factors of the coefficient of $x^{2}=3$

The possible rational roots are ±1, 1/3

Let, x – 1 = 0

=> x = 1

$f(1)=3(1)^{3}-(1)^{2}-3(1)+1$

= 3 – 1 – 3 + 1

= 0

So, x – 1 is the factor of f(x)

Now, divide f(x) with (x – 1) to get other factors

By long division method,

$3 x^{2}+2 x-1$

$x-13 x^{3}-x^{2}-3 x+1$

$3 x^{3}-x^{2}$

(-) $(+)$

$2 x^{2}-3 x$

$2 x^{2}-2 x$

 

$(-) \quad(+)$

- x + 1

- x + 1

(+)  (-)

0

$\Rightarrow 3 x^{3}-x^{2}-3 x+1=(x-1)\left(3 x^{2}+2 x-1\right)$

Now,

$3 x^{2}+2 x-1=3 x^{2}+3 x-x-1$

= 3x(x + 1) -1(x + 1)

= (3x – 1)(x + 1)

Hence, $3 x^{3}-x^{2}-3 x+1=(x-1)(3 x-1)(x+1)$