Finding Sum and Product of Zeros for a Cubic Polynomial

The question asks for the sum and product of the zeros of the given cubic polynomial \(p(y) = y^3 - 2y^2 - y + 2\). For a general cubic polynomial of the form \(ay^3 + by^2 + cy + d\), the relationships between the zeros (\(\alpha\), \(\beta\), \(\gamma\)) and the coefficients are given by Vieta's formulas:

• Sum of zeros: \(\alpha + \beta + \gamma = -\frac{b}{a}\)
• Sum of the product of zeros taken two at a time: \(\alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}\)
• Product of zeros: \(\alpha\beta\gamma = -\frac{d}{a}\)

In the given polynomial \(p(y) = y^3 - 2y^2 - y + 2\), we can identify the coefficients by comparing it to the standard form \(ay^3 + by^2 + cy + d\):

• \(a = 1\) (coefficient of \(y^3\))
• \(b = -2\) (coefficient of \(y^2\))
• \(c = -1\) (coefficient of \(y\))
• \(d = 2\) (constant term)

Calculating the Sum of Zeros

Using the formula for the sum of zeros:

\(\alpha + \beta + \gamma = -\frac{b}{a}\)

Substituting the values of \(a\) and \(b\):

\(\alpha + \beta + \gamma = -\frac{-2}{1}\)

\(\alpha + \beta + \gamma = \frac{2}{1}\)

\(\alpha + \beta + \gamma = 2\)

Calculating the Product of Zeros

Using the formula for the product of zeros:

\(\alpha\beta\gamma = -\frac{d}{a}\)

Substituting the values of \(a\) and \(d\):

\(\alpha\beta\gamma = -\frac{2}{1}\)

\(\alpha\beta\gamma = -2\)

Comparing with Options

Based on our calculations, the sum of the zeros is 2 and the product of the zeros is -2.

Let's examine the given options:

The values we calculated, \(\alpha\beta\gamma = -2\) and \(\alpha + \beta + \gamma = 2\), match the first option.