Show that (x – 2), (x + 3) and (x – 4) are the factors of

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)

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