Show that (x + 4), (x - 3) and (x - 7) are the factors of

Solution:

Here, $f(x)=x^{3}-6 x^{2}-19 x+84$

The factors given are (x + 4), (x - 3) and (x - 7)

To prove that g(x) is the factor of f(x),

The results of f(- 4) , f(3) and f(7) should be zero

Let, x + 4 = 0

⟹ x = - 4

Substitute the value of x in f(x)

$f(-4)=(-4)^{3}-6(-4)^{2}-19(-4)+84$

$=-64-\left(6^{*} 16\right)-\left(19^{*}(-4)\right)+84$

= - 64 - 96 + 76 + 84

= 160 - 160

= 0

Let, x - 3 = 0

⟹ x = 3

Substitute the value of x in f(x)

$f(3)=(3)^{3}-6(3)^{2}-19(3)+84$

= 27 – (6 * 9) – (19 * 3) + 84

= 27 – 54 – 57 + 84

= 111 – 111

= 0

Let, x - 7 = 0

⟹ x = 7

Substitute the value of x in f(x)

$f(7)=(7)^{3}-6(7)^{2}-19(7)+84$

= 343 – (6 * 49) – (19 * 7) + 84

= 343 – 294 - 133 + 84

= 427 – 427

= 0

Since, the results are 0 g(x) is the factor of f(x)