We are given the polynomial $f(x) = 2x^3 + x^2 - 5x + 2$ and the information that $x + 2$ is a factor.

Step 1: Apply the Factor Theorem

The Factor Theorem tells us that if $x + 2$ is a factor of the polynomial, then $x = -2$ must be a root of the polynomial. We can use synthetic division or polynomial division to divide $f(x)$ by $x + 2$.

Step 2: Perform synthetic division

We divide $f(x) = 2x^3 + x^2 - 5x + 2$ by $x + 2$. The coefficients of $f(x)$ are: $2, 1, -5, 2.$

Set up synthetic division using $x = -2$:

-2 & 2 & 1 & -5 & 2 \\ & & -4 & 6 & -2 \\ \hline & 2 & -3 & 1 & 0 \\ \end{array}$$ The remainder is 0, which confirms that $$x + 2$$ is indeed a factor. The quotient is $$2x^2 - 3x + 1$$. ### Step 3: Solve the quadratic equation Now we need to find the zeros of $$2x^2 - 3x + 1$$. We solve this quadratic equation using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$$ where $$a = 2$$, $$b = -3$$, and $$c = 1$$. First, calculate the discriminant: $$\Delta = (-3)^2 - 4(2)(1) = 9 - 8 = 1.$$ Now, apply the quadratic formula: $$x = \frac{-(-3) \pm \sqrt{1}}{2(2)} = \frac{3 \pm 1}{4}.$$ Thus, the two solutions are: $$x = \frac{3 + 1}{4} = \frac{4}{4} = 1,$$ and $$x = \frac{3 - 1}{4} = \frac{2}{4} = \frac{1}{2}.$$ ### Step 4: List all the real zeros The real zeros of the polynomial are $$x = -2$$, $$x = 1$$, and $$x = \frac{1}{2}$$. ### Final Answer: The real zeros of $$f(x) = 2x^3 + x^2 - 5x + 2$$ are: $$-2, 1, \frac{1}{2}.$$ Let me know if you'd like any further explanation or help!