Question

20 Points!!!!

Which real numbers are zeros of the function?

f(x)=3x^4-x^3-27x^2+9x

select each correct answer.

-3
-1
-1/3
0
1/3
1
3

Answer 1

The zeroes are -3, 0 , 1/3 and 3

Explanation:

The zeroes are -3, 0 , 1/3 and 3

Answer 2

The zeros of the given equation are
`0\ , (1)/(3)\ ,-3\ ,\ 3`.

Explanation:

Given f(x)=3x^4-x^3-27x^2+9x=0

Solution,

`f(x)=3x^4-x^3-27x^2+9x=0\\x^3*(3x-1)\ -9x*(3x-1)\ =0\\(3x-1)*(x^3-9x)\ =0`

For 1st zero we take;

`(3x-1) =0\\3x\ =1\\x\ =(1)/(3)`

Now for second zero;

`(x^3-9x)\ =0\\x*(x^2-9)\ =0\\ x\ =(0)/((x^2-9)) \ =0\\x\ =0`

Now for third and forth zeros;

`(x^3-9x)\ =0\\x*(x^2-9)\ =0\\(x^2-9)\ =0\\(x^2-3^2)\ =0\\(x+3)*(x-3)\ =0\\x\ =3\ or -3`

Hence all the zeros of the given equation are
`0\ , (1)/(3)\ ,-3\ ,\ 3`.