Cubic Functions and Equations

The sum is $-\dfrac{b}{a} = -\dfrac{-6}{1} = 6$. The roots are 1, 2 and 3, which do add to 6.

For a quadratic, sum $= -\dfrac{b}{a} = 7$ and product $= \dfrac{c}{a} = 12$. The roots are 3 and 4.

Product $= -\dfrac{d}{a} = -\dfrac{9}{3} = -3$.

$\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 25 - 12 = 13$. The roots are 2 and 3, and $4 + 9 = 13$ ✓

Sum $= 6$; pairs $= (-1)(2) + (2)(5) + (5)(-1) = -2 + 10 - 5 = 3$; product $= -10$. Then $x^3 - 6x^2 + 3x - (-10) = 0$.

Complex roots of a real polynomial come in conjugate pairs. With only three roots available, the remaining one must be real.