Determining the Number of Real Roots of a Cubic Polynomial using Sums and Products

Let \( \alpha, \beta \) and \( \gamma \) be the roots of the polynomial \( P(x) = 2x^3 + 4x^2 + 5x + 6 \).

Part 1

Find \( \alpha + \beta + \gamma \).

\( \alpha + \beta + \gamma = \displaystyle -\frac{4}{2} = -2 \)

Part 2

Find \( \alpha \beta + \beta \gamma + \alpha \gamma \).

\( \alpha \beta + \beta \gamma + \alpha \gamma = \displaystyle \frac{5}{2} \)

Part 3

Find \( \alpha \beta \gamma \).

\( \alpha \beta \gamma = \displaystyle – \frac{6}{2} = -3 \)

Part 4

Find \( \alpha^2 \beta \gamma + \alpha \beta^2 \gamma + \alpha \beta \gamma^2 \).

\( \begin{align} \alpha^2 \beta \gamma + \alpha \beta^2 \gamma + \alpha \beta \gamma^2 &= \alpha \beta \gamma (\alpha + \beta + \gamma) \\ &= -3 \times -2 \\ &= 6 \end{align} \)

Part 5

Find \( \alpha^2 + \beta^2 + \gamma^2 \).

\( \displaystyle \begin{align} \alpha^2 + \beta^2 + \gamma^2 &= (\alpha + \beta + \gamma)^2 – 2(\alpha \beta + \beta \gamma + \alpha \gamma) \\ &= (-2)^2 – 2 \times \frac{5}{2} \\ &= 4-5 \\ &= -1 \end{align} \)

Part 6

Determine how many of the roots (zeros) of \( P(x) \) are real.

Since \( P(x) \) has real coefficients, then any complex root occurs in conjugate pairs, in the form \( a \pm bi \).
Therefore \( P(x) \) has either 3 real roots or 1 real root and 2 complex roots in a conjugate pair.
As \( \alpha^2 + \beta^2 + \gamma^2 = -1 \lt 0 \), then \( \alpha, \beta \) and \( \gamma \) cannot be all real.
Thus \( P(s) \) must have 1 real root and a pair of complex conjugate roots.
That is, \( P(x) \) has only one real root.

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