Using factor theorem, factorize of the polynomials:

Solution:

Given, $f(x)=x^{4}-7 x^{3}+9 x^{2}+7 x-10$

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)

$f(x)=14-7(1)^{3}+9(1)^{2}+7(1)-10$

= 1 – 7 + 9 + 7 – 10

= 10 – 10

= 0

(x – 1) is the factor of f(x)

Similarly, the other factors are (x + 1), (x – 2), (x – 5)

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

So, f(x) = k(x – 1)(x + 1)(x – 2)(x – 5)

$\Rightarrow x^{4}-7 x^{3}+9 x^{2}+7 x-10=k(x-1)(x+1)(x-2)(x-5)$

Put x = 0 on both sides

0 – 0 + 0 – 10 = k(-1)(1)(-2)(-5)

– 10 = k(-10)

=> k = 1

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

f(x) = (1)(x – 1)(x + 1)(x – 2)(x – 5)

= (x – 1)(x + 1)(x – 2)(x – 5)

So, $x^{4}-7 x^{3}+9 x^{2}+7 x-10$

= (x – 1)(x + 1)(x – 2)(x – 5)