If f(x) = 6x^3 - 29x^2 -40x -12 and f(6) = 0, then find all the zeros of f(x) algebraically

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If f(x) = 6x^3 - 29x^2 -40x -12 and f(6) = 0, then find all the zeros of f(x) algebraically

11) obtuse

13)right

15) acute

[tex]State if each triangle is acute, right or obtuse[/tex]

the argument of alfred t. mahan, henry cabot lodge, and albert j. beveridge was the economically and mass production the the americans could not consume themselves. this led expansion to sell to other foreign nations for profits.

step-by-step explanation:

this is my explanation

[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]