Respuesta :
Answer:
The zeros are x+3 ,[tex]x-(3+\sqrt{5})[/tex] and [tex]x-(3-\sqrt{5})[/tex]
and is given by
[tex]f(x)=x^3-3x^2-14x+12=(x+3)(x-(3+\sqrt{5}))(x-(3-\sqrt{5}))[/tex]
Step-by-step explanation:
Given polynomial function is [tex]f(x)=x^3-3x^2-14x+12[/tex]
To find the zeros:
Equate the polynomial function to zero
That is f(x)=0
[tex]f(x)=x^3-3x^2-14x+12=0[/tex]
By using synthetic division method we can find zeros
-3_| 1 -3 -14 12
0 -3 18 -12
_______________
1 -6 4 0
Therefore x+3 is a zero
Ie., x=-3
Now the quadratic equation is [tex]x^2-6x+4=0[/tex]
Here a=1 , b=-6 and c=4
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(4)}}{2(1)}[/tex]
[tex]x=\frac{6\pm\sqrt{36-16}}{2}[/tex]
[tex]x=\frac{6\pm\sqrt{20}}{2}[/tex]
[tex]x=\frac{6\pm2\sqrt{5}}{2}[/tex]
[tex]x=3\pm\sqrt{5}[/tex]
Therefore [tex]x=3+\sqrt{5}[/tex] and [tex]x=3-\sqrt{5}[/tex]
[tex]x-(3+\sqrt{5})[/tex] and [tex]x-(3-\sqrt{5})[/tex]
Therefore the zeros are x+3 ,[tex]x-(3+\sqrt{5})[/tex] and [tex]x-(3-\sqrt{5})[/tex]
[tex]f(x)=x^3-3x^2-14x+12=(x+3)(x-(3+\sqrt{5}))(x-(3-\sqrt{5}))[/tex]
