We are given the cubic polynomial function:

$f(x) = 2x^3 - 12x^2 + 26x - 20$

Additionally, we are told that $x - 2x - 2$ is a factor of $f(x)$, which seems to indicate a factor of the form $x - c$ for some value of $c$.

Step 1: Factor out $(x - 2x - 2)$

First, let's simplify the expression for the factor you mentioned: $x - 2x - 2$.

It seems this might actually mean $(x - 2)$, which would be a linear factor of the polynomial.

Step 2: Use Synthetic Division to divide $f(x)$ by $x - 2$

We will divide $f(x) = 2x^3 - 12x^2 + 26x - 20$ by $x - 2$ using synthetic division to check if it is indeed a factor.

Synthetic Division Setup

Coefficients of $f(x)$: $2, -12, 26, -20$

2 & 2 & -12 & 26 & -20 \\ & & 4 & -16 & 20 \\ \hline & 2 & -8 & 10 & 0 \\ \end{array}$$ From the division, the quotient is $$2x^2 - 8x + 10$$, and the remainder is 0, confirming that $$x - 2$$ is indeed a factor of $$f(x)$$. ### Step 3: Solve the remaining quadratic equation Now, we need to solve for the zeros of the quadratic factor $$2x^2 - 8x + 10$$. The quadratic equation is: $$2x^2 - 8x + 10 = 0$$ #### Step 3.1: Apply the quadratic formula The quadratic formula is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ For the quadratic $$2x^2 - 8x + 10$$, the coefficients are: $$a = 2, \quad b = -8, \quad c = 10$$ Substituting these values into the quadratic formula: $$x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(2)(10)}}{2(2)}$$ $$x = \frac{8 \pm \sqrt{64 - 80}}{4}$$ $$x = \frac{8 \pm \sqrt{-16}}{4}$$ $$x = \frac{8 \pm 4i}{4}$$ $$x = 2 \pm i$$ ### Step 4: Find the zeros The zeros of the polynomial $$f(x)$$ are: $$x = 2 \quad \text{(from the factor } x - 2\text{)}$$ $$x = 2 + i \quad \text{(from the quadratic factor)}$$ $$x = 2 - i \quad \text{(from the quadratic factor)}$$ ### Final Answer: The zeros of $$f(x) = 2x^3 - 12x^2 + 26x - 20$$ are: $$x = 2, \quad x = 2 + i, \quad x = 2 - i$$ --- Would you like more detailed steps or explanations on any part of this solution? Also, here are 5 related questions you might find helpful: 1. How would you factor a cubic polynomial completely? 2. What is the significance of complex zeros in a polynomial? 3. Can synthetic division be used for higher-degree polynomials? 4. What are the possible types of zeros for a cubic function? 5. How can the quadratic formula be used to solve polynomials with complex solutions? **Tip:** Always check the remainder when using synthetic division to ensure the divisor is a true factor of the polynomial.