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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

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Answer:


6, (2)/(3),(1)/(2)

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
6x^3-29x^2-40x-12 and divisor
x-6:

quotient:
(6x^3)/(x)=6x^2

2. Multiply
x-6 by
6x ^ 2:
6x^3-36x^2.

Subtract
6x^3-36x^2 from
6x^3-29x^2-40x-12 to obtain a new remainder:
7x^2-40x-12

Now, divide the new remainder by
x-6. We repeat the previous steps:

3. quotient:
(7x^2)/(x)=7x

4. Multiply
x-6 by
7x:
7x^2+42x

Subtract
7x^2+42x from
7x^2-40x-12 to obtain a new remainder:
2x-12.

5. Divide
2x-12 by
x-6. We obtain
(2x-12)/(x-6)=2 and the new remainder is 0.

So, the quotiente
(6x^3-29x^2-40x-12)/(x-6)=6x^2+7x+2

Then,
6x^3-29x^2-40x-12=(x-6)(6x^2+7x+2)

For find the other two zeros of f(x) we have the zeros of
6x^2+7x+2.

We use the formula:
x_(1,2)=(-b\pm√(b^2-4ac))/(2a),

Then,
x_(1,2)=(7\pm√(7^2-4*6*2))/(2*6)=(7\pm 1)/(12)\\x_1=(8)/(12)=(2)/(3)\\x_2=(6)/(12)=(1)/(2)

Then the zeros of f(x) are:
6, (2)/(3),(1)/(2)

User Nijm
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