Show that $(x+4),(x-3)$ and $(x-7)$ are the factors of $x^{3}-6 x^{2}-19 x+84$
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)