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.