Answer:
Option C) is correct
That is the zeros of f(x) is 5,one third ,-one half
Step-by-step explanation:
Given function is [tex]f(x)=6x^3-29x^2-6x+5[/tex]
To find the zeros of the given function using synthetic division :
5_| 6 -29 -6 5
0 30 5 -5
___________________
6 1 -1 0
Therefore x=5 is a zero
The quadratic equation is [tex]6x^2+x-1=0[/tex]
[tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
Here a=6 ,b=1 and c=-1
[tex]x=\frac{-1\pm \sqrt{1^2-4(6)(-1)}}{2(6)}[/tex]
[tex]=\frac{-1\pm \sqrt{1+24}}{12}[/tex]
[tex]=\frac{-1\pm \sqrt{25}}{12}[/tex]
[tex]=\frac{-1\pm 5}{12}[/tex]
Therefore [tex]x=\frac{-1\pm 5}{12}[/tex]
[tex]x=\frac{-1+5}{12}[/tex] and [tex]x=\frac{-1-5}{12}[/tex]
[tex]x=\frac{4}{12}[/tex] and [tex]x=\frac{-6}{12}[/tex]
[tex]x=\frac{1}{3}[/tex] and [tex]x=\frac{-1}{2}[/tex]
Therefore the zeros are 5,[tex]\frac{1}{3}[/tex] and [tex]\frac{-1}{2}[/tex]
Therefore Option C) is correct
That is the zeros of f(x) is 5,one third ,-one half
