Question

Which of the following represents the zeros of f(x) = 6x3 − 29x2 − 6x + 5? (1 point)

Answer 1

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 $f(x)=6x^3-29x^2-6x+5$

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 $6x^2+x-1=0$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Here a=6 ,b=1 and c=-1

$x=\frac{-1\pm \sqrt{1^2-4(6)(-1)}}{2(6)}$

$=\frac{-1\pm \sqrt{1+24}}{12}$

$=\frac{-1\pm \sqrt{25}}{12}$

$=\frac{-1\pm 5}{12}$

Therefore $x=\frac{-1\pm 5}{12}$

$x=\frac{-1+5}{12}$ and $x=\frac{-1-5}{12}$

$x=\frac{4}{12}$ and $x=\frac{-6}{12}$

$x=\frac{1}{3}$ and $x=\frac{-1}{2}$

Therefore the zeros are 5,$\frac{1}{3}$ and $\frac{-1}{2}$

Therefore Option C) is correct

That is the zeros of f(x) is 5,one third ,-one half