Answer:
[tex]6, \frac{2}{3},\frac{1}{2}[/tex]
Step-by-step explanation:
Since 6 is a zero f f(x), then (x-6) is a factor of f(x). Now we will find the others factors
First, we divide f(x) by x-6:
1. divide the highest degree coefficients of the denominator [tex]6x^3-29x^2-40x-12[/tex] and divisor [tex]x-6[/tex]:
quotient: [tex]\frac{6x^3}{x}=6x^2[/tex]
2. Multiply [tex]x-6[/tex] by [tex]6x ^ 2[/tex]:[tex]6x^3-36x^2[/tex].
Subtract [tex]6x^3-36x^2[/tex] from [tex]6x^3-29x^2-40x-12[/tex] to obtain a new remainder: [tex]7x^2-40x-12[/tex]
Now, divide the new remainder by [tex]x-6[/tex]. We repeat the previous steps:
3. quotient: [tex]\frac{7x^2}{x}=7x[/tex]
4. Multiply [tex]x-6[/tex] by [tex]7x[/tex]:[tex]7x^2+42x[/tex]
Subtract [tex]7x^2+42x[/tex] from [tex]7x^2-40x-12[/tex] to obtain a new remainder: [tex]2x-12[/tex].
5. Divide [tex]2x-12[/tex] by [tex]x-6[/tex]. We obtain [tex]\frac{2x-12}{x-6}=2[/tex] and the new remainder is 0.
So, the quotiente [tex]\frac{6x^3-29x^2-40x-12}{x-6}=6x^2+7x+2[/tex]
Then, [tex]6x^3-29x^2-40x-12=(x-6)(6x^2+7x+2)[/tex]
For find the other two zeros of f(x) we have the zeros of [tex]6x^2+7x+2[/tex].
We use the formula: [tex]x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex],
Then, [tex]x_{1,2}=\frac{7\pm\sqrt{7^2-4*6*2}}{2*6}=\frac{7\pm 1}{12}\\x_1=\frac{8}{12}=\frac{2}{3}\\x_2=\frac{6}{12}=\frac{1}{2}[/tex]
Then the zeros of f(x) are: [tex]6, \frac{2}{3},\frac{1}{2}[/tex]
