Using factor theorem, factorize of the polynomials:

Let, $f(x)=x^{3}-6 x^{2}+3 x+10$

The constant term in f(x) is 10

The factors of 10 are ± 1, ± 2, ± 5, ± 10

Let, x + 1 = 0

=> x = -1

Substitute the value of x in f(x)

$f(-1)=(-1)^{3}-6(-1)^{2}+3(-1)+10$

= -1 – 6 – 3 + 10

= 0

Similarly, the other factors (x – 2) and (x – 5) of f(x)

Since, f(x) is a polynomial having a degree 3, it cannot have more than three linear factors.

∴ f(x) = k(x + 1)(x – 2)(x – 5)

Substitute x = 0 on both sides

$=>x^{3}-6 x^{2}+3 x+10=k(x+1)(x-2)(x-5)$

=> 0 – 0 + 0 + 10 = k(1)(-2)(-5)

=> 10 = k(10)

=> k = 1

Substitute k = 1 in f(x) = k(x + 1)(x – 2)(x – 5)

f(x) = (1)(x + 1)(x – 2)(x – 5 )

so, $x^{3}-6 x^{2}+3 x+10=(x+1)(x-2)(x-5)$