Question

find all zeros of the function (f(x) = 2x^3 - 14x - 12). enter the zeros separated by commas.

Explanation:

🆕 New Concept Discovered: Rational Zeros Theorem
Finding possible rational roots of a polynomial

Step 1: Simplify the function

To find the zeros of \( f(x) = 2x^3 - 14x - 12 \), we set the function equal to zero:

$$ 2x^3 - 14x - 12 = 0 $$

We can simplify this equation by dividing all terms by their greatest common factor, \( 2 \):

$$ x^3 - 7x - 6 = 0 $$

Step 2: Find a rational zero

According to the Rational Zeros Theorem, any rational zero of the polynomial must be a factor of the constant term, \( -6 \), divided by a factor of the leading coefficient, \( 1 \).

The factors of \( -6 \) are:

$$ \pm 1, \pm 2, \pm 3, \pm 6 $$

Let's test these possible factors in our simplified equation \( g(x) = x^3 - 7x - 6 \) to find one that equals zero:

• Test \( x = 1 \):

$$ (1)^3 - 7(1) - 6 = 1 - 7 - 6 = -12 eq 0 $$

• Test \( x = -1 \):

$$ (-1)^3 - 7(-1) - 6 = -1 + 7 - 6 = 0 $$

Since \( x = -1 \) makes the equation equal to zero, \( x = -1 \) is a zero, and \( (x + 1) \) is a factor.

Step 3: Divide the polynomial

Now, we divide \( x^3 - 7x - 6 \) by \( (x + 1) \) using synthetic division or polynomial long division to find the remaining quadratic factor:

Using synthetic division with \( -1 \):

 -1 |   1    0   -7   -6
    |       -1    1    6
    --------------------
        1   -1   -6    0

The quotient is the quadratic expression:

$$ x^2 - x - 6 $$

Step 4: Solve the remaining quadratic equation

Set the quadratic quotient to zero to find the remaining zeros:

$$ x^2 - x - 6 = 0 $$

Factor the quadratic expression:

$$ (x - 3)(x + 2) = 0 $$

This gives the remaining zeros:

$$ x = 3 \quad \text{and} \quad x = -2 $$

Combining all the found values, the zeros of the function are \( -1 \), \( -2 \), and \( 3 \).

Answer:

-1, -2, 3