To show that $c = -2$ and $c = \frac{1}{2}$ are zeros of $P(x) = 2x^4 - x^3 - 14x^2 - 5x + 6$, we will evaluate $P(x)$ at $x = -2$ and $x = \frac{1}{2}$ step by step, and then use the Factor Theorem. Afterward, we will find the remaining zeros.

Step 1: Check if $c = -2$ is a zero of $P(x)$.

We need to evaluate $P(-2)$:

$P(x) = 2x^4 - x^3 - 14x^2 - 5x + 6$

$P(-2) = 2(-2)^4 - (-2)^3 - 14(-2)^2 - 5(-2) + 6$

Calculating each term:

$(-2)^4 = 16, \quad (-2)^3 = -8, \quad (-2)^2 = 4$

$P(-2) = 2(16) - (-8) - 14(4) - 5(-2) + 6$ $P(-2) = 32 + 8 - 56 + 10 + 6$ $P(-2) = 0$

Thus, $P(-2) = 0$. By the Factor Theorem, this means $x + 2$ is a factor of $P(x)$, so $-2$ is a zero of $P(x)$.

Step 2: Check if $c = \frac{1}{2}$ is a zero of $P(x)$.

Now we evaluate $P\left(\frac{1}{2}\right)$:

$P\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^4 - \left(\frac{1}{2}\right)^3 - 14\left(\frac{1}{2}\right)^2 - 5\left(\frac{1}{2}\right) + 6$

Calculating each term:

$\left(\frac{1}{2}\right)^4 = \frac{1}{16}, \quad \left(\frac{1}{2}\right)^3 = \frac{1}{8}, \quad \left(\frac{1}{2}\right)^2 = \frac{1}{4}$

$P\left(\frac{1}{2}\right) = 2\left(\frac{1}{16}\right) - \frac{1}{8} - 14\left(\frac{1}{4}\right) - 5\left(\frac{1}{2}\right) + 6$ $P\left(\frac{1}{2}\right) = \frac{2}{16} - \frac{1}{8} - \frac{14}{4} - \frac{5}{2} + 6$ $P\left(\frac{1}{2}\right) = \frac{1}{8} - \frac{1}{8} - \frac{35}{2} + 6$ $P\left(\frac{1}{2}\right) = 0$

Thus, $P\left(\frac{1}{2}\right) = 0$. By the Factor Theorem, this means $x - \frac{1}{2}$ is a factor of $P(x)$, so $\frac{1}{2}$ is a zero of $P(x)$.

Step 3: Factor $P(x)$ and find all other zeros.

Since $x = -2$ and $x = \frac{1}{2}$ are zeros, $P(x)$ can be factored as:

$P(x) = (x + 2)\left(x - \frac{1}{2}\right)Q(x)$

To find the remaining zeros, we can perform polynomial division or factorization to find $Q(x)$, the remaining quadratic factor.

The complete factored form of $P(x)$ is:

$P(x) = (x + 2)\left(2x - 1\right)(x^2 - 3)$

Thus, the remaining zeros come from solving $x^2 - 3 = 0$:

$x^2 = 3 \quad \Rightarrow \quad x = \pm \sqrt{3}$

Final Answer:

The zeros of $P(x)$ are: $x = -2, \frac{1}{2}, \sqrt{3}, -\sqrt{3}$

Would you like further details or clarifications? Here are some related questions:

1. How do you perform synthetic division to find polynomial factors?
2. What is the Factor Theorem and how does it work?
3. How do you solve higher-degree polynomials?
4. How can you find zeros of polynomials graphically?
5. What role does the discriminant play in finding zeros?

Tip: Always verify the potential zeros by substituting them back into the polynomial.