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

Question

asked Jul 11, 2020 88.2k views

2 Answers

Damon Abdiel · Answer 1 · 2020-07-15T13:11:06+0000

Answer:

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
$0\ , (1)/(3)\ ,-3\ ,\ 3$ .