Using factor theorem, factorize of the polynomials:

Solution:

Given,

$f(x)=x^{4}-2 x^{3}-7 x^{2}+8 x+12$

The constant term f(x) is equal is 12

The factors of 12 are ± 1, ± 2, ± 3, ± 4, ± 6, ± 12

Let, x + 1 = 0

=> x = -1

Substitute the value of x in f(x)

$f(-1)=(-1)^{4}-2(-1)^{3}-7(-1)^{2}+8(-1)+12$

= 1 + 2 – 7 – 8 + 12

= 0

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

Similarly, (x + 2), (x – 2), (x – 3) are also the factors of f(x)

Since, f(x) is a polynomial of degree 4, it cannot have more than four linear factors.

=> f(x) = k(x + 1)(x + 2)(x – 3)(x – 2)

$=>x^{4}-2 x^{3}-7 x^{2}+8 x+12=k(x+1)(x+2)(x-3)(x-2)$

Substitute x = 0 on both sides,

=> 0 – 0 – 0 + 12 = k(1)(2)(- 2)(- 3)

=> 12 = 12K

=> k = 1

Substitute k = 1 in f(x) = k(x – 2)(x + 1)(x + 2)(x – 3)

f(x) = (x – 2)(x + 1)(x + 2)(x – 3)

so, $x^{4}-2 x^{3}-7 x^{2}+8 x+12=(x-2)(x+1)(x+2)(x-3)$