Question:
Show that $(x-2),(x+3)$ and $(x-4)$ are the factors of $x^{3}-3 x^{2}-10 x+24$
Solution:
Here, $f(x)=x^{3}-3 x^{2}-10 x+24$
The factors given are (x - 2), (x + 3) and (x - 4)
To prove that g(x) is the factor of f(x),
The results of f(2), f(-3) and f(4) should be zero
Let, x - 2 = 0
⟹ x = 2
Substitute the value of x in f(x)
$f(2)=2^{3}-3(2)^{2}-10(2)+24$
= 8 - (3 * 4) - 20 + 24
= 8 - 12 - 20 + 24
= 32 - 32
= 0
Let, x + 3 = 0
⟹ x = -3
Substitute the value of x in f(x)
$f(-3)=(-3)^{3}-3(-3)^{2}-10(-3)+24$
= -27 - 3(9) + 30 + 24
= -27 - 27 + 30 + 24
= 54 - 54
= 0
Let, x - 4 = 0
⟹ x = 4
Substitute the value of x in f(x)
$f(4)=(4)^{3}-3(4)^{2}-10(4)+24$
= 64 - (3 * 16) - 40 + 24
= 64 - 48 - 40 + 24
= 84 - 84
= 0
Since, the results are 0 g(x) is the factor of f(x)