we have the function
f(x)=x^3-5x^2-2x+24
Remember that
If x=3 is a zero of the given function
then
(x-3) is a factor of the given function
so
Divide the given function f(x) by the factor
x^3-5x^2-2x+24 : (x-3)
x^2-2x-8
-x^3+3x^2
--------------------------
-2x^2-2x+24
2x^2-6x
----------------------
-8x+24
8x-24
------------
0
therefore
x^3-5x^2-2x+24=(x-3)( x^2-2x-8)
Solve the quadratic equation
x^2-2x-8=0
using the formula
a=1
b=-2
c=-8
substitute
![x=(-(-2)\pm√(-2^2-4(1)(-8)))/(2(1))](https://img.qammunity.org/2023/formulas/mathematics/college/y1trg5t6b5ngskcgn7ncvfhjq6zzf335sz.png)
![x=(2\pm6)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/y99ayvg6345725bmaz4irc2103c05py4u5.png)
The values of x are
x=4 and x=-2
therefore
x^3-5x^2-2x+24=(x-3)(x+2)(x-4)
The zeros of the function f(x) are
x=3
x=4
x=-2