Question

find all zeros, real and non-real, of (f(x) = x^3 - 8x^2 + 14x + 68). enter the zeros separated by commas.

Explanation:

🆕 New Concept Discovered: Finding Zeros of Polynomial Functions
Finding where a polynomial equals zero using division and quadratic methods.

Step 1: Find a rational zero

To find the first zero of the cubic polynomial \( f(x) = x^3 - 8x^2 + 14x + 68 \), we test factors of the constant term, \( 68 \). The integer factors of \( 68 \) are \( \pm 1, \pm 2, \pm 4, \pm 17, \pm 34, \pm 68 \).

Let's test \( x = -2 \):

$$ f(-2) = (-2)^3 - 8(-2)^2 + 14(-2) + 68 $$

$$ f(-2) = -8 - 8(4) - 28 + 68 $$

$$ f(-2) = -8 - 32 - 28 + 68 $$

$$ f(-2) = -68 + 68 = 0 $$

Since \( f(-2) = 0 \), \( x = -2 \) is a zero, which means \( (x + 2) \) is a factor of the polynomial.

Step 2: Divide the polynomial

We use synthetic division to divide \( x^3 - 8x^2 + 14x + 68 \) by \( (x + 2) \) to find the remaining quadratic factor:

 -2 |   1   -8   14   68
    |       -2   20  -68
    --------------------
        1  -10   34    0

The quotient is the quadratic expression:

$$ x^2 - 10x + 34 $$

Step 3: Solve the quadratic factor

To find the remaining zeros, we set the quadratic factor to zero and solve using the quadratic formula:

$$ x^2 - 10x + 34 = 0 $$

Using \( a = 1 \), \( b = -10 \), and \( c = 34 \):

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

$$ x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(34)}}{2(1)} $$

$$ x = \frac{10 \pm \sqrt{100 - 136}}{2} $$

$$ x = \frac{10 \pm \sqrt{-36}}{2} $$

Since \( \sqrt{-36} = 6i \):

$$ x = \frac{10 \pm 6i}{2} $$

$$ x = 5 \pm 3i $$

The remaining complex zeros are \( 5 + 3i \) and \( 5 - 3i \).

Answer:

-2, 5 + 3i, 5 - 3i