Using factor theorem, factorize of the polynomials:

Solution:

Given,

$f(x)=x^{3}-10 x^{2}-53 x-42$

The constant in f(x) is - 42

The factors of - 42 are ± 1, ± 2, ± 3, ± 6, ± 7, ± 14, ± 21,± 42

Let, x + 1 = 0

=> x = - 1

$f(-1)=(-1)^{3}-10(-1)^{2}-53(-1)-42$

= -1 – 10 + 53 – 42

= 0

So., (x + 1) is the factor of f(x)

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

By long division,

$x^{2}-11 x-42$

$x+1 x^{3}-10 x^{2}-53 x-42$

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

(-)     (-)

$-11 x^{2}-53 x$

 

$-11 x^{2}-11 x$

(+)       (+)

- 42x – 42

- 42x – 42

(+)       (+)

0

$\Rightarrow x^{3}-10 x^{2}-53 x-42=(x+1)\left(x^{2}-11 x-42\right)$

Now,

$x^{2}-11 x-42=x^{2}-14 x+3 x-42$

= x(x – 14) + 3(x – 14)

= (x + 3)(x – 14)

Hence, $x^{3}-10 x^{2}-53 x-42=(x+1)(x+3)(x-14)$