We are tasked with finding all possible zeros of the polynomial function:

$P(x) = 2x^4 - 8x^3 + 4x^2 + 8x - 6$

Step 1: Apply the Rational Root Theorem

The Rational Root Theorem suggests that any rational solution, $p/q$, must be a factor of the constant term (here, -6) divided by a factor of the leading coefficient (here, 2).

• Factors of $-6$ (constant term): $\pm 1, \pm 2, \pm 3, \pm 6$
• Factors of $2$ (leading coefficient): $\pm 1, \pm 2$

Possible rational roots ( $\frac{p}{q}$ ) are:

$\pm 1, \pm \frac{1}{2}, \pm 2, \pm 3, \pm \frac{3}{2}, \pm 6$

Step 2: Test possible rational roots

We will substitute these values into $P(x)$ to check which ones yield $P(x) = 0$.

Test $x = 1$:

$P(1) = 2(1)^4 - 8(1)^3 + 4(1)^2 + 8(1) - 6 = 2 - 8 + 4 + 8 - 6 = 0$ So, $x = 1$ is a root.

Step 3: Perform synthetic division

Now that $x = 1$ is a root, we can use synthetic division to divide $P(x)$ by $(x - 1)$.

Synthetic division of $P(x)$ by $(x - 1)$ results in:

$2x^3 - 6x^2 - 2x + 6$

Step 4: Factor the cubic polynomial

Now, factor $2x^3 - 6x^2 - 2x + 6$. We can take out a common factor of 2:

$2(x^3 - 3x^2 - x + 3)$

We can apply synthetic division again or use grouping to factor $x^3 - 3x^2 - x + 3$. Testing for possible rational roots shows that $x = 3$ is a root.

Step 5: Divide again

Synthetic division of $x^3 - 3x^2 - x + 3$ by $(x - 3)$ results in:

$x^2 - 1 = (x - 1)(x + 1)$

Step 6: Write the complete factorization

So, the factored form of $P(x)$ is:

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

Step 7: Find the zeros

The zeros of the polynomial are:

$x = 1 \quad (\text{with multiplicity 2}), \quad x = -1, \quad x = 3$

Thus, the possible zeros are:

$x = 1, -1, 3$

Would you like further explanation on any step or details? Here are five questions related to polynomial functions and factorization:

1. How do you check if a number is a zero of a polynomial?
2. What is the Rational Root Theorem and when do you use it?
3. How does synthetic division help simplify polynomials?
4. What is the significance of multiplicity in roots?
5. How can grouping be used to factor cubic polynomials?

Tip: Always test simpler roots first (like $\pm 1$) when applying the Rational Root Theorem for easier calculations.