To find all the zeros of the polynomial function $p(x) = 2x^3 - 5x^2 + x + 2$, we will follow these steps:

Step 1: Use the Rational Root Theorem

The Rational Root Theorem helps us identify possible rational roots of the polynomial. According to the theorem, any rational root $\frac{p}{q}$ of a polynomial is a divisor of the constant term divided by a divisor of the leading coefficient.

• The constant term is $2$, so the divisors of 2 are $\pm 1, \pm 2$.
• The leading coefficient is $2$, so the divisors of 2 are $\pm 1, \pm 2$.

The possible rational roots are: $\pm 1, \pm 2, \pm \frac{1}{2}$

Step 2: Test the possible rational roots

We will now substitute the possible values into the polynomial and check which one gives $p(x) = 0$.

• For $x = 1$: $p(1) = 2(1)^3 - 5(1)^2 + (1) + 2 = 2 - 5 + 1 + 2 = 0$ Therefore, $x = 1$ is a root.

Step 3: Perform synthetic division

Since $x = 1$ is a root, we can divide $p(x)$ by $x - 1$ using synthetic division to simplify the polynomial.

Set up the synthetic division:

1 & 2 & -5 & 1 & 2 \\ & & 2 & -3 & -2 \\ \hline & 2 & -3 & -2 & 0 \\ \end{array}$$ The quotient is $$2x^2 - 3x - 2$$. ### Step 4: Factor the quadratic Now, we need to solve the quadratic $$2x^2 - 3x - 2$$. We can factor this quadratic: $$2x^2 - 3x - 2 = (2x + 1)(x - 2)$$ ### Step 5: Solve for the remaining roots Set each factor equal to 0 and solve for $$x$$: 1. $$2x + 1 = 0$$ gives $$x = -\frac{1}{2}$$. 2. $$x - 2 = 0$$ gives $$x = 2$$. ### Final Answer: The zeros of the polynomial $$p(x) = 2x^3 - 5x^2 + x + 2$$ are: $$x = 1, \quad x = -\frac{1}{2}, \quad x = 2$$ Would you like a detailed explanation of any step? Here are five related questions to consider: 1. How do we use synthetic division to simplify a polynomial? 2. Can all polynomials be factored using the Rational Root Theorem? 3. What happens if none of the rational roots work? 4. How does the quadratic formula apply if factoring is difficult? 5. What is the role of complex numbers when finding roots? **Tip:** Always verify your roots by substituting them back into the original polynomial equation.