1)Making the substitution

t = x - \frac{1}{2}

, the cubic equation

2 t^{3} + 3 t^{2} - 11 t - 6 = 0

becomes

t^{3} + \frac{3}{2} t^{2} - \frac{11}{2} t - 3 = (x - \frac{1}{2})^{3} + \frac{3}{2} (x - \frac{1}{2})^{2} - \frac{11}{2} (x - \frac{1}{2}) - 3 = 0 .

Simplifying this equation we get:

x^{3} - \frac{3}{2} x^{2} + \frac{3}{4} x - \frac{1}{8} + \frac{3}{2} (x^{2} - x + \frac{1}{4}) - \frac{11}{2} (x - \frac{1}{2}) - 3 = 0

and collecting like terms we get

x^{3} - \frac{25}{4} x = 0

which has solutions

x = 0

and

\pm \frac{5}{2}

. The equivalent values of

t

are

0 - \frac{1}{2} = - \frac{1}{2}

and

\pm \frac{5}{2} - \frac{1}{2} = 2

and -3.

(4) The equation $Q (x) = x^{3} + px + q = 0$ has derivative $Q' (x) = 3 x^{2} + p$ and turning points where $x = \pm \sqrt{-} \frac{p}{3}$ . Write $p = - s^{2}$ then the turning points are $x = \pm \frac{s}{\sqrt{3}}$ . The condition for three real roots is that one turning point has a positive $y$ coordinate and the other a negative $y$ coordinate, i.e.

$[\frac{s^{3}}{3 \sqrt{3}} - \frac{s^{3}}{\sqrt{3}} + q] [- \frac{s^{3}}{3 \sqrt{3}} + \frac{s^{3}}{\sqrt{3}} + q] \leq 0$

which reduces to

$[q - \frac{2 s^{3}}{3 \sqrt{3}}] [q + \frac{2 s^{3}}{3 \sqrt{3}}] \leq 0$

i.e.

$q^{2} \leq \frac{4 s^{6}}{27}, q^{2} \leq \frac{- 4 p^{3}}{27} .$

cubic equation regions