The polynomial given in the image is:

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

To find the possible zeros (also called roots) of this polynomial, we can follow several methods. Here is the process:

1. Use the Rational Root Theorem

The Rational Root Theorem helps us identify possible rational roots by examining factors of the constant term and the leading coefficient.

• The constant term is $-6$, so the factors of $-6$ are: $\pm 1, \pm 2, \pm 3, \pm 6$
• The leading coefficient is $2$, so the factors of $2$ are: $\pm 1, \pm 2$

Using the Rational Root Theorem, the possible rational roots are all ratios of the factors of the constant term over the factors of the leading coefficient: $\pm \frac{1}{1}, \pm \frac{2}{1}, \pm \frac{3}{1}, \pm \frac{6}{1}, \pm \frac{1}{2}, \pm \frac{3}{2}$ So, the possible rational zeros are: $\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}$

2. Test the possible rational roots

We now substitute these values into the polynomial to check if any satisfy $P(x) = 0$.

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

3. Factor the polynomial

Now that we know $x = 1$ is a root, we can factor the polynomial using synthetic division or polynomial division. Dividing $P(x)$ by $x - 1$ gives us a quotient, which we can further analyze to find other zeros.

4. Solve the remaining quadratic

After factoring and reducing the polynomial, the remaining terms can be factored or solved using the quadratic formula or other methods.

Do you need help with performing the synthetic division or further factoring the polynomial?

Expanding on the concept, consider these questions:

1. How does the Rational Root Theorem help identify possible zeros?
2. What happens if none of the rational roots work in a polynomial?
3. How do we apply synthetic division to factor a polynomial?
4. Can complex numbers also be solutions to this polynomial?
5. What are the differences between rational and irrational roots?

Tip: When testing possible roots, always start with simpler values (like $\pm 1$) to reduce the work needed for polynomial factoring.