1)Making the substitution t = x - 1/2, the cubic equation
2t3+ 3t2 - 11t - 6 = 0
becomes
t3+ 3
2
 t2 - 11
2
 t - 3 = (x - 1
2
 )3 + 3
2
 (x - 1
2
 )2 - 11
2
 (x - 1
2
 ) - 3 = 0.
Simplifying this equation we get:
x3 - 3
2
 x2 + 3
4
 x - 1
8
 + 3
2
 (x2-x+ 1
4
 ) - 11
2
 (x - 1
2
 ) - 3 = 0
and collecting like terms we get
x3 - 25
4
 x = 0
which has solutions x=0 and ±5/2. The equivalent values of t are 0-1/2 = -1/2 and ±5/2 - 1/2 = 2 and -3.

(4) The equation Q(x)=x3 + px +q = 0 has derivative Q¢(x) = 3x2 + p and turning points where x = ±Ö-p/3. Write p=-s2 then the turning points are
x=± s
Ö3

. 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.
[ s3
3Ö3
 - s3
Ö3
 + q][- s3
3Ö3
 + s3
Ö3
 + q] £ 0
which reduces to
[q - 2s3
3Ö3
 ][q + 2s3
3Ö3
 ] £ 0
i.e.
q2 £ 4s6
27
 ,    q2 £ -4p3
27
 .
cubic equation regions